ACM算法模板

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acm 算法模板适合初学者使用

三角形面积计算 (1)字典树模板 (2)求线段所在直线 (5)求外接圆 (5)求内接圆 (6)判断点是否在直线上 (8)简单多边形面积计算公式 (8)stein算法求最大共约数 (9)最长递增子序列模板——o(nlogn算法实现) (9)判断图中同一直线的点的最大数量 (10)公因数和公倍数 (12)已知先序中序求后序 (12)深度优先搜索模板 (13)匈牙利算法——二部图匹配BFS实现 (15)带输出路径的prime算法 (17)prime模板 (18)kruskal模板 (19)dijsktra (22)并查集模板 (23)高精度模板 (24)三角形面积计算//已知三条边和外接圆半径，公式为s = a*b*c/(4*R)double GetArea(double a, double b, double c, double R){return a*b*c/4/R;}//已知三条边和内接圆半径，公式为s = prdouble GetArea(double a, double b, double c, double r){return r*(a+b+c)/2;}//已知三角形三条边，求面积double GetArea(doule a, double b, double c){double p = (a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c));}//已知道三角形三个顶点的坐标struct Point{double x, y;Point(double a = 0, double b = 0){x = a; y = b;}};double GetArea(Point p1, Point p2, Point p3){double t =-p2.x*p1.y+p3.x*p1.y+p1.x*p2.y-p3.x*p2.y-p1.x*p3.y+p2.x*p3.y;if(t < 0) t = -t;return t/2;}字典树模板#include <stdio.h>#include <string.h>#include <memory.h>#define BASE_LETTER 'a'#define MAX_TREE 35000#define MAX_BRANCH 26struct{int next[MAX_BRANCH]; //记录分支的位置int c[MAX_BRANCH]; //查看分支的个数int flag; //是否存在以该结点为终止结点的东东，可以更改为任意的属性}trie[MAX_TREE];int now;void init(){now = 0;memset(&trie[now], 0, sizeof(trie[now]));now ++;}int add (){memset(&trie[now], 0, sizeof(trie[now]));return now++;}int insert( char *str){int pre = 0, addr;while( *str != 0 ){addr = *str - BASE_LETTER;if( !trie[pre].next[addr] )trie[pre].next[addr] = add();trie[pre].c[addr]++;pre = trie[pre].next[addr];str ++;}trie[pre].flag = 1;return pre;}int search( char *str ){int pre = 0, addr;while( *str != 0 ){addr = *str - BASE_LETTER;if ( !trie[pre].next[addr] )return 0;pre = trie[pre].next[addr];str ++;}if( !trie[pre].flag )return 0;return pre;}pku2001题，源代码：void check( char *str ){int pre = 0, addr;while(*str != 0){addr = *str - BASE_LETTER;if( trie[pre].c[addr] == 1) {printf("%c\n", *str);return;}printf("%c", *str);pre = trie[pre].next[addr];str ++;}printf("\n");}char input[1001][25];int main(){int i = 0,j;init();while(scanf("%s", input[i]) != EOF){getchar();insert(input[i]);i++;}for(j = 0; j < i; j ++){printf("%s ", input[j]);check(input[j]);}return 0;}求线段所在直线//*****************************线段所在的直线struct Line{double a, b, c;};struct Point{double x, y;}Line GetLine(Point p1, Point p2){//ax+by+c = 0返回直线的参数Line line;line.a = p2.y - p1.y;line.b = p1.x - p2.x;line.c = p2.x*p1.y - p1.x*p2.y;return line;}求外接圆//***************已知三角形三个顶点坐标，求外接圆的半径和坐标********************struct Point{double x, y;Point(double a = 0, double b = 0){x = a; y = b;}};struct TCircle{double r;Point p;}double distance(Point p1, Point p2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}double GetArea(doule a, double b, double c){double p = (a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c));}TCircle GetTCircle(Point p1, Point p2, Point p3){double a, b, c;double xa,ya, xb, yb, xc, yc, c1, c2;TCircle tc;a = distance(p1, p2);b = distance(p2, p3);c = distance(p3, p1);//求半径tc.r = a*b*c/4/GetArea(a, b, c);//求坐标xa = p1.x; ya = p1.b;xb = p2.x; yb = p2.b;xc = p3.x; yc = p3.b;c1 = (xa*xa + ya*ya - xb*xb - yb*yb)/2;c2 = (xa*xa + ya*ya - xc*xc - yc*yc)/2;tc.p.x = (c1*(ya-yc) - c2*(ya-yb))/((xa-xb)*(ya-yc) - (xa-xc)*(ya-yb)); tc.p.y = (c1*(xa-xc) - c2*(xa-xb))/((ya-yb)*(xa-xc) - (ya-yc)*(xa-xb));return tc;}求内接圆struct Point{double x, y;Point(double a = 0, double b = 0){x = a; y = b;}};struct TCircle{double r;Point p;}double distance(Point p1, Point p2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}double GetArea(doule a, double b, double c){double p = (a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c));}TCircle GetTCircle(Point p1, Point p2, Point p3){double a, b, c;double xa,ya, xb, yb, xc, yc, c1, c2, f1, f2;double A,B,C;TCircle tc;a = distance(p1, p2);b = distance(p3, p2);c = distance(p3, p1);//求半径tc.r = 2*GetArea(a, b, c)/(a+b+c);//求坐标A = acos((b*b+c*c-a*a)/(2*b*c));B = acos((a*a+c*c-b*b)/(2*a*c));C = acos((a*a+b*b-c*c)/(2*a*b));p = sin(A/2); p2 = sin(B/2); p3 = sin(C/2);xb = p1.x; yb = p1.b;xc = p2.x; yc = p2.b;xa = p3.x; ya = p3.b;f1 = ( (tc.r/p2)*(tc.r/p2) - (tc.r/p)*(tc.r/p) + xa*xa - xb*xb + ya*ya - yb*yb)/2;f2 = ( (tc.r/p3)*(tc.r/p3) - (tc.r/p)*(tc.r/p) + xa*xa - xc*xc + ya*ya - yc*yc)/2;tc.p.x = (f1*(ya-yc) - f2*(ya-yb))/((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb)); tc.p.y = (f1*(xa-xc) - f2*(xa-xb))/((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));return tc;}判断点是否在直线上//**************判断点是否在直线上********************* //判断点p是否在直线[p1,p2]struct Point{double x,y;};bool isPointOnSegment(Point p1, Point p2, Point p0){//叉积是否为0，判断是否在同一直线上if((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y) != 0)return false;//判断是否在线段上if((p0.x > p1.x && p0.x > p2.x) || (p0.x < p1.x && p0.x < p2.x)) return false;if((p0.y > p1.y && p0.y > p1.y) || (p0.y < p1.y && p0.y < p2.y)) return false;return true;}简单多边形面积计算公式struct Point{double x, y;Point(double a = 0, double b = 0){x = a; y = b;}};Point pp[10];double GetArea(Point *pp, int n){//n为点的个数，pp中记录的是点的坐标int i = 1;double t = 0;for(; i <= n-1; i++)t += pp[i-1].x*pp[i].y - pp[i].x*pp[i-1].y;t += pp[n-1].x*pp[0].y - pp[0].x*pp[n-1].y;if(t < 0) t = -t;return t/2;}stein算法求最大共约数int gcd(int a,int b){if (a == 0) return b;if (b == 0) return a;if (a % 2 == 0 && b % 2 == 0) return 2 * gcd(a/2,b/2); else if (a % 2 == 0) return gcd(a/2,b);else if (b % 2 == 0) return gcd(a,b/2);else return gcd(abs(a-b),min(a,b));}最长递增子序列模板——o(nlogn算法实现)#include <stdio.h>#define MAX 40000int array[MAX], B[MAX];int main(){int count,i,n,left,mid,right,Blen=0,num;scanf("%d",&count); //case的个数while(count--){scanf("%d",&n); //每组成员的数量Blen = 0;for(i=1;i<=n;i++)scanf("%d",&array[i]); //读入每个成员for(i=1;i<=n;i++){num = array[i];left = 1;right = Blen;while(left<=right){mid = (left+right)/2;if(B[mid]<num)left = mid+1;elseright = mid-1;}B[left] = num;if(Blen<left)Blen++;}printf("%d\n",Blen);//输出结果}return 1;}判断图中同一直线的点的最大数量#include <iostream>#include <cstdio>#include <memory>using namespace std;#define MAX 1010 //最大点的个数struct point{int x,y;}num[MAX];int used[MAX][MAX*2]; //条件中点的左边不会大于1000,just equal MAX int countN[MAX][MAX*2];#define abs(a) (a>0?a:(-a))int GCD(int x, int y){int temp;if(x < y){temp = x; x = y; y = temp;}while(y != 0){temp = y;y = x % y;x = temp;}return x;}int main(){int n,i,j;int a,b,d,ans;while(scanf("%d", &n)==1){//initeans = 1;memset(used, 0, sizeof(used));memset(countN, 0, sizeof(countN));//readfor(i = 0; i < n; i++)scanf("%d%d", &num[i].x, &num[i].y);for(i = 0; i < n-1; i++){for(j = i+1; j < n; j++){b = num[j].y-num[i].y;a = num[j].x-num[i].x;if(a < 0) //这样可以让（2，3）（－2，－3）等价{a = -a; b = -b;}d = GCD(a,abs(b));a /= d;b /= d; b += 1000;//条件中点的左边不会大于1000if(used[a][b] != i+1){used[a][b] = i+1;countN[a][b] = 1;}else{countN[a][b]++;if(ans < countN[a][b])ans = countN[a][b];}}//for}//forprintf("%d\n", ans+1);}return 0;}公因数和公倍数int GCD(int x, int y){int temp;if(x < y){temp = x; x = y; y = temp;}while(y != 0){temp = y;y = x % y;x = temp;}return x;}int beishu(int x, int y){return x * y / GCD(x,y);}已知先序中序求后序#include <iostream>#include <string>using namespace std;string post;void fun(string pre, string mid){if(pre == "" || mid == "") return;int i = mid.find(pre[0]);fun(pre.substr(1,i), mid.substr(0,i));fun(pre.substr(i+1, (int)pre.length()-i-1), mid.substr(i+1, (int)mid.length()-i-1));post += pre[0];}int main(){string pre, mid;while(cin >> pre){cin >> mid;post.erase();fun(pre, mid);cout << post << endl;}return 0;}深度优先搜索模板int t; //t用来标识要搜索的元素int count; //count用来标识搜索元素的个数int data[m][n]; //data用来存储数据的数组//注意，数组默认是按照1……n存储，即没有第0行//下面是4个方向的搜索,void search(int x, int y){data[x][y] = *; //搜索过进行标记if(x-1 >= 1 && data[x-1][y] == t){count++;search(x-1,y);}if(x+1 <= n && data[x+1][y] == t){count++;search(x+1,y);}if(y-1 >= 1 && data[x][y-1] == t){count++;search(x,y-1);}if(y+1 <= n && data[x][y+1] == t){count++;search(x,y+1);}}//下面是8个方向的搜索void search(int x, int y){data[x][y] = *; //搜索过进行标记if(x-1 >= 1){if(data[x-1][y] == t){count++;search(x-1,y);}if(y-1 >= 1 && data[x-1][y-1] == t) {count++;search(x-1,y-1);}if(y+1 <= n && data[x-1][y+1] == t) {count++;search(x-1,y+1);}}if(x+1 <= n){if(data[x+1][y] == t){count++;search(x+1,y);}if(y-1 >= 1 && data[x+1][y-1] == t) {count++;search(x+1,y-1);}if(y+1 <= n && data[x+1][y+1] == t) {count++;search(x+1,y+1);}}if(y-1 >= 1 && data[x][y-1] == t){count++;search(x,y-1);}if(y+1 <= n && data[x][y+1] == t){count++;search(x,y+1);}}匈牙利算法——二部图匹配BFS实现//匈牙利算法实现#define MAX 310 //二部图一侧顶点的最大个数int n,m; //二分图的两个集合分别含有n和m个元素。

ACM 算法模板2

目录一、图论 (2)1.1.最小生成树类prim算法 (2)1.2.拓扑排序 (4)1.3.最短源路径Folyd实现 (5)1.4.关键路径实现算法 (6)1.5.二分图最大匹配的匈牙利算法 (8)1.6.并查集 (9)二、动规 (11)2.1.求最长子序列 (11)2.2.求解最长升序列长度及子序列 (12)2.3.完全背包问题 (13)2.4.0-1背包问题 (14)2.5.母函数DP算法求组合数 (15)2.6.滚动数组求回文串问题 (17)三、贪心 (18)3.1.时间安排问题 (18)3.2.求最大子段和 (19)3.3.贪心求最少非递减序列数 (20)四、数论 (21)4.1.简单求Cnk问题 (21)4.2.巧求阶乘位数 (21)4.3.线性算法求素数 (22)五、其他 (22)5.1.采用位操作递归求解示例 (22)5.2.Stack和Queue用法 (23)5.3.map使用详解 (24)5.4.字典树建立与查找 (25)5.5.KMP匹配算法 (26)5.6.后缀数组求最长连续公共子序列长度 (28)5.7.循环字符串最小位置表示及同构判断 (30)5.8.求哈夫曼树编码长度 (32)5.9.堆排序算法 (34)5.10.线段树着色问题 (35)六、附： (39)6.1.C++最值常量 (39)6.2.类型转换 (39)6.3.String常用函数举例 (39)6.4.C++常用头文件 (40)一、图论1.1.最小生成树类prim算法1.1.1下标从1开始#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define SIZE 101#define MAXSIZE 10201int n,nline;/*n 个点,nline行关系*/int in[SIZE];struct Point{int x,y;/*编号从1开始*/int v;/*根据实际情况更改类型*/}p[MAXSIZE];/*n*(n-1)/2*/int cmp(Point a,Point b){return a.v<b.v;}int prim(){int dis,count,i,j;memset(in,0,sizeof(in));in[p[1].x]=in[p[1].y]=1;dis=p[1].v;count=n-1;while(count--){/*做n-1次*/for(j=2;j<nline;j++){if((in[p[j].x]&&!in[p[j].y])||(!in[p[j].x]&&in[p[j].y])){in[p[j].x]=1;in[p[j].y]=1;dis+=p[j].v;break;}}}return dis;}int main(){int x,y,v,i;while(scanf("%d",&n)&&n){if(n==1){/*有可能输入的为1个点*/printf("0\n");continue;}nline=n*(n-1)/2+1;for(i=1;i<nline;i++){scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].v);}sort(p+1,p+nline,cmp);printf("%d\n",prim());}return 0;}1.1.2下标从0开始#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define SIZE 101#define MAXSIZE 10201int n,nline;/*n 个点,nline行关系*/int in[SIZE];struct Point{int x,y;/*编号从1开始*/int v;/*根据实际情况更改类型*/}p[MAXSIZE];/*n*(n-1)/2*/int cmp(Point a,Point b){return a.v<b.v;}int prim(){int dis,count,i,j;memset(in,0,sizeof(in));in[p[0].x]=in[p[0].y]=1;dis=p[0].v;count=n-1;while(count--){/*做n-1次*/for(j=1;j<nline;j++){if((in[p[j].x]&&!in[p[j].y])||(!in[p[j].x]&&in[p[j].y])){in[p[j].x]=1;in[p[j].y]=1;dis+=p[j].v;break;}}}return dis;}int main(){int x,y,v,i;while(scanf("%d",&n)&&n){if(n==1){/*有可能输入的为1个点*/printf("0\n");continue;}nline=n*(n-1)/2;for(i=0;i<nline;i++){scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].v);}sort(p,p+nline,cmp);printf("%d\n",prim());}return 0;}1.2.拓扑排序#include<iostream>#include<cstdio>#include<cstring>#define M 501using namespace std;int map[M][M],degree[M];int ne;/*个数*/void topo(){int i,j,k;for(i=0;i<ne;i++){j=1;while(j<=ne&&degree[j])j++;//直到一个度为零的顶点,这里不检查有多个度为零的情况//if(j>ne){break;}不是拓扑结构if(i)printf(" ");printf("%d",j);degree[j]=-1;for(k=1;k<=ne;k++){degree[k]-=map[j][k];}}printf("\n");}int main(){int a,b,i,j,nline;/*nline行*/while(scanf("%d%d",&ne,&nline)!=EOF){memset(map,0,sizeof(map));memset(degree,0,sizeof(degree));while(nline--){scanf("%d%d",&a,&b);map[a][b]=1;/*a to b*/}for(i=1;i<=ne;i++){for(j=1;j<=ne;j++){if(map[i][j])degree[j]++;}}topo();/*拓扑*/}return 0;}1.3.最短源路径Folyd实现#include<iostream>#include<cstdio>#include<cstring>#define M 201using namespace std;int n,map[M][M],start,end;void folyd(){int i,j,k;for(i=0;i<n;i++){for(j=0;j<n;j++){for(k=0;k<n;k++){if(map[j][i]==-1||map[i][k]==-1)continue;if(map[j][k]==-1||map[j][i]+map[i][k]<map[j][k]){map[j][k]=map[j][i]+map[i][k];}}}}}int main(){int nline,i,j,a,b,v;while(scanf("%d%d",&n,&nline)!=EOF){memset(map,-1,sizeof(map));for(i=0;i<n;i++){map[i][i]=0;}for(i=0;i<nline;i++){scanf("%d%d%d",&a,&b,&v);/*编号从0开始*/if(map[a][b]==-1||map[a][b]>v){//一个点到另一个有多条路map[b][a]=map[a][b]=v;}}scanf("%d%d",&start,&end);folyd();if(map[start][end]!=-1){printf("%d\n",map[start][end]);}else printf("-1\n");}return 0;}1.4.关键路径实现算法#include<iostream>#include<cstdio>#include<cstring>#define M 501using namespace std;int map[M][M],degree[M],dp[M];int ne;/*个数*/int topoplus(){int i,j,k,maxnum;for(i=0;i<ne;i++){j=1;while(j<=ne&&degree[j])j++;//直到一个度为零的顶点,这里不检查有多个度为零的情况//if(j>ne){break;}不是拓扑结构degree[j]=-1;for(k=1;k<=ne;k++){if(map[j][k]){if(map[j][k]+dp[j]>dp[k]){dp[k]=map[j][k]+dp[j];}degree[k]--;}}}for(i=0;i<=ne;i++){if(dp[i]>maxnum){maxnum=dp[i];}}return maxnum;}int main(){int a,b,v,i,j,nline;/*nline行*/while(scanf("%d%d",&ne,&nline)!=EOF){memset(map,0,sizeof(map));memset(degree,0,sizeof(degree));memset(dp,0,sizeof(dp));while(nline--){scanf("%d%d%d",&a,&b,&v);map[a][b]=v;/*a to b，v>0*/}for(i=1;i<=ne;i++){for(j=1;j<=ne;j++){if(map[i][j])degree[j]++;}}printf("%d\n",topoplus());/*拓扑改进*/}return 0;}1.5.二分图最大匹配的匈牙利算法#include<iostream>#include<cstdio>#define N 301using namespace std;int isuse[N]; //记录y中节点是否使用int lk[N]; //记录当前与y节点相连的x的节点int mat[N][N];//记录连接x和y的边，如果i和j之间有边则为1，否则为0 int gn,gm; //二分图中x和y中点的数目int can(int t){int i;for(i=1;i<=gm;i++){//下标从1开始if(isuse[i]==0 && mat[t][i]){isuse[i]=1;if(lk[i]==-1 || can(lk[i])){lk[i]=t;return 1;}}}return 0;}int MaxMatch(){int i,num=0;memset(lk,-1,sizeof(lk));for(i=1;i<=gn;i++){memset(isuse,0,sizeof(isuse));if(can(i))num++;}return num;}int main(){int t,i,j,k,tmp;scanf("%d",&t);while(t--){scanf("%d%d",&gn,&gm);memset(mat,0,sizeof(mat));//主要得到mat这个数组for(i=1;i<=gn;i++){scanf("%d",&k);for(j=1;j<=k;j++){scanf("%d",&tmp);mat[i][tmp]=1;//注意从1开始}}if(MaxMatch()==gn){printf("YES\n");}else printf("NO\n");}return 0;}/*In:23 33 1 2 32 1 21 13 32 1 32 1 31 1Out:YESNO*/1.6.并查集#include<iostream>#include<cstdio>using namespace std;const int N=1010;int pre[N];void Merge(int x,int y){int i,t,rx=x,ry=y;while(pre[rx]!=-1)//搜索x的树根rx=pre[rx];while(pre[ry]!=-1)//搜索y的树根ry=pre[ry];i=x;//压缩xwhile(pre[i]!=-1){t=pre[i];pre[i]=rx;i=t;}i=y;//压缩ywhile(pre[i]!=-1){t=pre[i];pre[i]=rx;i=t;}if(ry!=rx)//合并pre[ry]=rx;return;}int main(){int x,y,i,ans,n,m;while(scanf("%d",&n)&&n){scanf("%d",&m);memset(pre,-1,sizeof(pre));for(i=0;i<m;i++){//x与y连通scanf("%d %d",&x,&y);Merge(x,y);}ans=0;for(i=1;i<=n;i++)if(pre[i]==-1)ans++;printf("%d\n",ans-1);}}/*/showproblem.php?pid=1232 in:4 21 34 3999 0out:1998*/二、动规2.1.求最长子序列#include<iostream>#include<cstdio>#define M 1001using namespace std;char a[M],b[M];int dp[M+1][M+1],lena,lenb;void init(){int i,j;for(i=0;i<=lena;i++){for(j=0;j<=lenb;j++){dp[i][j]=0;}}}int cmax(int x,int y){return x>y?x:y;}int main(){int i,j,len;while(scanf("%s",a)!=EOF){scanf("%s",b);init();//下面这步很重要，否则会超时lena=strlen(a);lenb=strlen(b);for(i=0;i<lena;i++){for(j=0;j<lenb;j++){if(a[i]==b[j]){dp[i+1][j+1]=dp[i][j]+1;}else{dp[i+1][j+1]=cmax(dp[i][j+1],dp[i+1][j]);}}}//子序列长度printf("%d\n",dp[i][j]);/*打印出子序列*/len=1;for(i=1;i<=lena;i++){for(j=1;j<=lenb;j++){if(len==dp[i][j]){printf("%c",a[i-1]);len++;break;}}}printf("\n");}return 0;}2.2.求解最长升序列长度及子序列#include<iostream>#include<cstdio>#define M 1001using namespace std;int a[M],dp[M];void init(int n){int i;for(i=0;i<n;i++){dp[i]=1;}}int lis(int n){int i,j,maxlen=1;//初始长度为1for(i=n-2;i>=0;i--){for(j=n-1;j>i;j--){if(a[i]<a[j]){if(dp[j]+1>dp[i]){dp[i]=dp[j]+1;}if(dp[i]>maxlen)maxlen=dp[i];}}}return maxlen;}void showlis(int n,int maxlen){int i;for(i=0;i<n;i++){if(dp[i]==maxlen){printf("%d ",a[i]);maxlen--;}}printf("\n");}int main(){int t,n,i,maxlen;scanf("%d",&t);while(t--){/*1<=n<=1000*/scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);}init(n);maxlen=lis(n);printf("%d\n",maxlen);//显示最长升序列showlis(n,maxlen);}return 0;}2.3.完全背包问题#include<iostream>#include<cstring>#include<cstdio>using namespace std;int type[]={150,200,350};//种类int dp[10001];int max(int a,int b){return a>b?a:b;}int main(){int t,n,i,j;scanf("%d",&t);while(t--){scanf("%d",&n);memset(dp,0,sizeof(dp));for(i=0;i<3;i++){for(j=type[i];j<=n;j++){//剩余容量为j时装的东西量最大dp[j]=max(dp[j],dp[j-type[i]]+type[i]);}}printf("%d\n",n-dp[n]);}return 0;}2.4.0-1背包问题#include<iostream>#include<cstdio>#include<cstring>#define M 1002using namespace std;int val[M],wei[M],dp[M][M];int cmax(int a,int b){return a>b?a:b;}int main(){int t,n,w,i,j;scanf("%d",&t);while(t--){scanf("%d%d",&n,&w);for(i=1;i<=n;i++){scanf("%d",&val[i]);}for(i=1;i<=n;i++){scanf("%d",&wei[i]);}memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){for(j=0;j<=w;j++){if(j>=wei[i])dp[i][j]=cmax(dp[i-1][j-wei[i]]+val[i],dp[i-1][j]);else dp[i][j]=dp[i-1][j];}}printf("%d\n",dp[n][w]);}return 0;}2.5.母函数DP算法求组合数2.5.1.求母函数各系数值DP#include <iostream>#define M 17#define MAX 305using namespace std;int c1[MAX],c2[MAX],add[M+1];//add[]保存M种类void init(){int i;for(i=1;i<=M;i++){add[i]=i*i;}}int solve(int n){int i,j,k;//c1[k],c2[k]表示展开式中x^k的系数memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));c1[0]=c2[0]=1;//使用前i种币时的情况，也即母函数展开前i个多项式的乘积 for(i=1;i<=M;i++){//求新的多项式中的系数for(j=0;j<n;j++){for(k=1;j+k*add[i]<=n;k++){c2[j+k*add[i]]+=c1[j];}}for(k=0;k<=n;k++){//滚动数组c1[k]=c2[k];}}return c1[n];}int main(){int n;init();while(scanf("%d",&n)&&n){printf("%d\n",solve(n));}return 0;}2.5.2.状态继承类DP求某个和是否存在#include<cstdio>#include<iostream>#include<cstring>using namespace std;bool dp[250002];int val[101],num[101];//对应的值和数量int main(){int n,i,j,sum,k;while(scanf("%d",&n)&&n>0){sum=0;for(i=0;i<n;i++){scanf("%d%d",&val[i],&num[i]);sum+=val[i]*num[i];}//dp中保存所有可能的组合memset(dp,0,sizeof(dp));dp[0]=1;//遍历n种物品for(i=0;i<n;i++){//对区间求for(j=sum/2;j>=0;j--){if(!dp[j]){for(k=1;k<=num[i]&&k*val[i]<=j;k++){dp[j]|=dp[j-k*val[i]];}}}}for(i=sum/2;i>=0;i--){if(dp[i]){printf("%d %d\n",sum-i,i);break;}}}return 0;}/*in:210 120 1320 230 1-1out:20 1040 40*/2.6.滚动数组求回文串问题#include<cstdio>#include<iostream>#include<cstring>#define M 5001using namespace std;char str[M],rstr[M];int dp[2][M];//滚动DPint cmax(int x,int y){return x>y?x:y;}int main(){int n,i,j,s1,s2;while(scanf("%d",&n)!=EOF){scanf(" %s",str);//反转字符数组for(i=0;i<n;i++){rstr[n-i-1]=str[i];}rstr[n]='\0';memset(dp,0,sizeof(dp));for(i=0;i<n;i++){for(j=0;j<n;j++){s1=i%2;s2=(i+1)%2;if(str[i]==rstr[j]){dp[s1][j+1]=dp[s2][j]+1;}else{dp[s1][j+1]=cmax(dp[s2][j+1],dp[s1][j]);}}}printf("%d\n",n-dp[(n-1)%2][n]);return 0;}/*/showproblem.php?pid=1513 in:5Ab3bdout:2*/三、贪心3.1.时间安排问题#include<iostream>#include<cstdio>#include<algorithm>#define M 101using namespace std;int n;struct Point{int s,e;}p[M];int cmp(Point a,Point b){if(a.s==b.s)return a.e<b.e;else return a.s<b.s;}int arrange(){int start,end,i,count=1;start=p[0].s;end=p[0].e;for(i=1;i<n;i++){if(p[i].s>=start&&p[i].e<=end){start=p[i].s;end=p[i].e;}else if(p[i].s>=end){count++;end=p[i].e;}}return count;}int main(){int i;while(scanf("%d",&n)&&n){for(i=0;i<n;i++){scanf("%d%d",&p[i].s,&p[i].e);}sort(p,p+n,cmp);printf("%d\n",arrange());}return 0;}3.2.求最大子段和#include<iostream>using namespace std;int main(){int t,n,i,a[100002];int beg,end,x,y,cursum,maxsum;cin>>t;while(t--){cin>>n;for(i=0;i<n;i++){cin>>a[i];}beg=end=1;cursum=maxsum=a[0];x=y=1;for(i=1;i<n;i++){if(a[i]+cursum<a[i]){cursum=a[i];x=i+1;}else{cursum+=a[i];}if(cursum>maxsum){maxsum=cursum;beg=x;end=i+1;}}cout<<maxsum<<" "<<beg<<" "<<end<<endl;}return 0;}/*in:25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5out:14 1 47 1 6*/3.3.贪心求最少非递减序列数#include<iostream>#include<cstdio>#include<cstring>using namespace std;int a[1001];int main(){int n,i,j,len,count,high;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++){scanf("%d",&a[i]);}count=0;len=n;while(len){count++;high=30005;//最高值for(i=0;i<n;i++){if(a[i]&&a[i]<high){high=a[i];a[i]=0;//标记已用值len--;}}}printf("%d\n",count);}return 0;}/*in:8 389 207 155 300 299 170 158 65out:2*/四、数论4.1.简单求Cnk问题#include<iostream>using namespace std;int main(){int n,k,i;double sum;while(cin>>n>>k){if(n==0&&k==0)break;if(k>n-k)k=n-k;sum=1;for(i=1;i<=k;i++){sum*=(double)(n-k+i)/i*1.000000000001;//必需要乘 }cout<<(int)sum<<endl;}return 0;}4.2.巧求阶乘位数#include<iostream>#include<cmath>using namespace std;const double pi=acos(-1.0);//NOTES:piconst double e=2.71828182845904523536028747135266249775724709369995957; int main(){long long n,tt;cin>>tt;while (tt--){cin>>n;long long ans=(long long)((double)log10(sqrt(2*pi*n))+n*log10(n/e))+1;cout<<ans<<endl;}return 0;}4.3.线性算法求素数const int MAX=10000000;//求[2,MAX]间的素数bool isprime[MAX+1];int prime[MAX];//保存素数//返回素数表元素总数int getprime(){int i,j,pnum=0;//memset(isprime,0,sizeof(isprime));for(i=2;i<=MAX;i++){if(!isprime[i])prime[pnum++]=i;for(j=0;j<pnum&&prime[j]*i<=MAX;j++){isprime[prime[j]*i]=1;if(i%prime[j]==0)break;}}return pnum;}五、其他5.1.采用位操作递归求解示例#include<iostream>#include<cstdio>using namespace std;unsigned short in[50001],ste;/*用16位的ste保存16种状态*/ unsigned short power[]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768}; int n,m,maxnum,i,j;void dfs(int s,int count){if(count>maxnum)maxnum=count;for(i=s;i<m;i++){if(!(ste&in[i])){/*如果相与为0，说明材料未被使用*/ ste=ste|in[i];dfs(i+1,count+1);ste=ste&(~in[i]);}}}int main(){int tn,t;while(scanf("%d%d",&n,&m)!=EOF){if(n==0||m==0){printf("0\n");continue;}for(i=0;i<m;i++){scanf("%d",&tn);in[i]=0;for(j=1;j<=tn;j++){scanf("%d",&t);/*材料编号降为从0开始，防止益处*/in[i]=in[i]|power[t-1];}}ste=0;maxnum=0;dfs(0,0);printf("%d\n",maxnum);}return 0;}5.2.Stack和Queue用法#include<iostream>#include<stack>#include<queue>using namespace std;int main(){stack<int> s;queue<int> q;int a[]={1,2,3,4};/*加入*/for(int i=0;i<4;i++){s.push(a[i]);q.push(a[i]);}/*读取stack*/cout<<"stack-size:"<<s.size()<<endl;for(int i=0;i<4;i++){cout<<s.top()<<" ";s.pop();}cout<<endl<<"queue-size:"<<q.size()<<endl;/*读取queue*/cout<<"front:"<<q.front()<<"back:"<<q.back()<<endl;for(int i=0;i<4;i++){cout<<q.front()<<" ";q.pop();}cout<<endl;return 0;}5.3.map使用详解#include<iostream>#include<map>#include<string>#include<iterator>using namespace std;int main(){map<string,int>m;map<string,int>::iterator p;map<string,int>::reverse_iterator q;m["bd"]=2;m["ba"]=1;m["aa"]=3;m["bd"]=4;//按从小到大遍历for(p=m.begin();p!=m.end();p++){//注意不能使用p<m.end() cout<<p->first<<" "<<p->second<<endl;}//按从大到小遍历for(q=m.rbegin();q!=m.rend();q++){cout<<q->first<<" "<<q->second<<endl;}m.erase(m.begin());cout<<m.size()<<endl;//清楚全部m.clear();cout<<m.empty()<<endl;return 0;}5.4.字典树建立与查找#include<iostream>#include<cstring>#include<cstdio>#define M 26using namespace std;int ii;//只在Tree中使用struct Tree{Tree* next[M];int val;Tree(){for(ii=0;ii<M;ii++){next[ii]=0;}val=0;}~Tree(){for(ii=0;ii<M;ii++){delete(next[ii]);}}};int main(){char word[20];int len,i,j,count;Tree* root=new Tree;Tree* p;//建立字典树过程while(gets(word)){if(strcmp(word,"")==0)break;len=strlen(word);p=root;for(i=0;i<len;i++){j=word[i]-'a';if(p->next[j]==0){p->next[j]=new Tree;}p=p->next[j];(p->val)++;}word[0]='\0';}while(scanf("%s",word)!=EOF){len=strlen(word);p=root;for(i=0;i<len;i++){j=word[i]-'a';if(p->next[j]!=0){p=p->next[j];}else break;}if(i==len)printf("%d\n",p->val);else printf("0\n");}return 0;}/*In:bananabandbeeabsoluteacmbabbandabcout:231*/5.5.KMP匹配算法#include<iostream>#include<cstdio>#include<cstring>#define M 10001using namespace std;int s[M*100],t[M],next[M];//得到next数组，下标均从1开始void getnext(int m){int i=1,j=0;next[1]=0;while(i<=m){if(j==0||t[i]==t[j]){++i;++j;next[i]=j;}else j=next[j];}}//找不到则返回-1int kmp(int n,int m){int i=0,j=1;getnext(m);while(i<=n&&j<=m){if(!j||s[i]==t[j]){++i;++j;}elsej=next[j];}if(j>m)return i-m;else return -1;}int main(){int test,m,n,i;scanf("%d",&test);while(test--){scanf("%d %d",&n,&m);//主串s,长度为nfor(i=1;i<=n;i++)scanf("%d",&s[i]);//横式串t,长度为mfor(i=1;i<=m;i++)scanf("%d",&t[i]);printf("%d\n",kmp(n,m));}return 0;}5.6.后缀数组求最长连续公共子序列长度#include<iostream>#include<cstdio>#include<algorithm>#define M 100001using namespace std;char message[M*2];/*后缀数组*/int height[M*2];int _array[2][M*2];int _rank[2][M*2];int cnt[M*2];int *array, *rank, *narray, *nrank;/*得到最长连续公共子序列长度*/int suffix(int len1,int len2,int len){int i,k;memset(cnt,0,1024);for(i=0;i<len;++i){++cnt[message[i]];}for(i=1;i<= 'z';++i){cnt[i]+=cnt[i-1];}array = _array[0];rank = _rank[0];for(i=len-1;i>=0;--i){array[--cnt[message[i]]]=i;}rank[array[0]] = 0;for(i=1;i<len;i++){rank[array[i]]=rank[array[i-1]];if(message[array[i]]!=message[array[i-1]]){ rank[array[i]]++;}}narray = _array[1];nrank = _rank[1];for(k=1;k<len&&rank[array[len-1]]<len-1;k<<=1){for(i=0;i<len;++i){cnt[rank[array[i]]]=i+1;}for(i=len-1;i>=0;--i){if(array[i] >= k){// array[i]是当前的最大值，所以array[i] - k//是其相同前缀中（rank相同）的最大值narray[--cnt[rank[array[i]-k]]]=array[i]-k;}}for(i=len-k;i<len;++i){//这些没有k后缀，所以他们是最后面的k个，他的位置已经比较出来narray[--cnt[rank[i]]]=i;}nrank[narray[0]] = 0;for (i=1;i<len;++i){nrank[narray[i]]=nrank[narray[i-1]];if(rank[narray[i]]!= rank[narray[i-1]] ||rank[narray[i]+k]!=rank[narray[i-1]+k]){//如果前缀的排名不同，则++；如果前缀相同，但是后缀不同，也++ nrank[narray[i]]++;}}swap(nrank, rank);swap(narray, array);}int ret=0,hei;for(i=1;i<len;++i){if(((array[i]<len1&&array[i-1]>=len1) ||(array[i]>=len1&&array[i-1]<len1))){hei = 0;while(message[array[i]+ hei]==message[array[i-1]+hei]){hei++;}ret = max(ret, hei);}}return ret;}int main(){int len,len1,len2;while(gets(message)!=NULL){len1 = strlen(message);gets(message + len1);len2 = strlen(message+len1);len = len1 + len2;printf("%d\n",suffix(len1,len2,len));}return 0;}5.7.循环字符串最小位置表示及同构判断#include<cstdio>#include<iostream>#include<cstring>#include<string>using namespace std;//返回两个字符串是否同构//len为s1或S2的长度，s1与s2等长//pos1与pos2为字符循环最小表示位置，从0开始bool CircularMatch(string s1, string s2, int len, int& pos1, int& pos2) {int p1 = 0, p2 = 0, k, t1, t2;pos1 = pos2 = -1;while (1) {k = 0;while (1) {t1 = (p1+k)%len; t2 = (p2+k)%len;if(s1[t1] > s2[t2]) {p1 = p1+k+1;if (p1 >= len) return false;break;}else if (s1[t1] < s2[t2]) {p2 = p2+k+1;if (p2 >= len) return false;break;}else k++;if (k == len) {pos1 = p1; pos2 = p2;return true;}}}}//返回字符串循环最小表示的位置,从0开始int MinCircularDenote(string s, int len) {int p1 = 0, p2 = 1, k, t1, t2;while (1) {k = 0;while (1) {t1 = (p1+k)%len; t2 = (p2+k)%len;if(s[t1] > s[t2]) {if (p1+k+1 <= p2) p1 = p2+1;else p1 = p1+k+1;if (p1 >= len) return p2;break;}else if (s[t1] < s[t2]) {if (p2+k+1 <= p1) p2 = p1+1;else p2 = p2+k+1;if (p2 >= len) return p1;break;}else k++;if (k == len)return (p1<p2 ? p1 : p2);}}}//返回字符串循环最大表示的位置,从0开始int MaxCircularDenote(string str,int len) {int p1 = 0, p2 = 1, k, t1, t2;while (1) {k = 0;while (1) {t1 = (p1+k)%len; t2 = (p2+k)%len;if(str[t1] < str[t2]) {if (p1+k+1 <= p2) p1 = p2+1; else p1 = p1+k+1;if (p1 >= len) return p2;break;}else if (str[t1] > str[t2]) {if (p2+k+1 <= p1) p2 = p1+1;else p2 = p2+k+1;if (p2 >= len) return p1;break;}else k++;if (k == len)return (p1<p2 ? p1 : p2);}}}int main(){string s1,s2;int pos1,pos2,len;while(cin>>s1>>s2){len=s1.length();cout<<MinCircularDenote(s1,len)<<endl;cout<<MaxCircularDenote(s1,len)<<endl;s1+=s1;//字符串加倍s2+=s2;cout<<CircularMatch(s1,s2,len, pos1,pos2)<<endl;cout<<pos1<<endl<<pos2<<endl;}return 0;}5.8.求哈夫曼树编码长度#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#define M 27using namespace std;char line[100001];int tree[M],tmp[M];//保存权值int cmp(int a,int b){return a>b;}//对序号为index统计长度int huffman(int n,int index){if(n==1)return 1;//一个点时返回1int i,j,count,max,len,t;for(i=0;i<n;i++){tmp[i]=tree[i];}count=max=n-1;len=0;while(count--){if(index==max||index==max-1){len++;index=max-1;}tmp[max-1]+=tmp[max];j=--max;while(j>0&&tmp[j]>tmp[j-1]){t=tmp[j];tmp[j]=tmp[j-1];tmp[j-1]=t;if(index==j){index--;}else if(index==j-1){index++;}j--;}}return len;}int main(){int i,j,len,sum,n;int ch[M];while(gets(line)!=NULL){if(strcmp(line,"END")==0)break;len=strlen(line);memset(ch,0,sizeof(ch));for(i=0;i<len;i++){if(line[i]=='_'){ch[0]++;}else ch[line[i]-'A'+1]++;}//权值压缩到tree[]数组中for(j=0,i=0;i<M;i++){if(ch[i]){tree[j++]=ch[i];}}n=j;sort(tree,tree+n,cmp);for(sum=0,i=0;i<n;i++){sum+=tree[i]*huffman(n,i);}printf("%d %d %.1lf\n",len*8,sum,len*8.0/(sum*1.0));}return 0;}/*/showproblem.php?pid=1053in:AAAAABCDTHE_CAT_IN_THE_HATENDout:64 13 4.9144 51 2.8*/5.9.堆排序算法#include<cstdio>#include<cstdlib>using namespace std;void adjust(int a[],int s,int len){int t,i;t=a[s];for(i=s*2;i<len;i*=2){if(i<(len-1)&&a[i]<a[i+1])i++;if(t>=a[i])break;a[s]=a[i];s=i;}a[s]=t;}void sort(int a[],int len){int i,t;for(i=(len-1)/2;i>=0;i--){adjust(a,i,len);}for(i=len-1;i>1;i--){a[i]=a[0];a[0]=t;adjust(a,0,i-1);}if(a[0]>a[1]){t=a[0];a[0]=a[1];a[1]=t;}}int main(){int a[1000],i,n;scanf("%d",&n);//for(i=0;i<n;i++)// scanf("%d",a+i);for(i=0;i<n;i++)a[i]=rand()%10000;sort(a,n);for(i=0;i<n;i++)printf("%d ",a[i]);printf("\n");return 0;}//2 38 4 99 10 2 22 1 -43 22 33 91 78 335.10.线段树着色问题#include <iostream>#define N 8003#define NoCol -1#define MulCol -2using namespace std;struct SegTree{int l,r,c;}st[N*4];//一般大小开成节点数的4倍，不需要担心空间问题int seg[N],col[N];int n,sat,end;//创建线段树void segTreeCre(int l,int r,int i=1){int mid;st[i].l=l;。

ACM模版

大数除法16进制#include <stdio.h>#include <string.h>#define max 200int i;char jinzhi[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'}; int chu(char a[],char b[]){int shan, yu;for(yu = i = 0; a[i]; i++){yu = yu * 10 + a[i] - '0';shan = yu / 16;yu = yu % 16;b[i] = shan + '0';}b[i] = 0;return yu;}int judge(char a[]){for(i = 0; a[i]; i++)if(a[i] != '0')return 0;return 1;}void f(char a[]){int newbig;if(!judge(a)){char b[max];newbig = chu(a,b);f(b);printf("%c",jinzhi[newbig]);}}int main(void){char bigint[max];while(scanf("%s", bigint) == 1){if(strlen(bigint) == 1 ){if(bigint[0] == '0')return 0;}f(bigint);printf("\n");memset(bigint,'0',sizeof(bigint));}return 0;}最大公约数int GCD(int num1,int num2)//最大公约数{if ( num1 % num2 == 0){return num2;}elsereturn GCD( num2,num1 % num2) ;}最小公倍数int LCM(int a,int b)//最小公倍数{int temp_lcm;temp_lcm=a*b/GCD(a,b); //最小公倍数等于两数之积除以最大公约数return temp_lcm;}//求至少多少天相遇，同时出发，至少多少长方形堆正方体，分元宝类问题都是运用最小公倍数最小生成树#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int inf = ( 1 << 20 ) ;int p[27]; // 并查集，用于判断两点是否直接或间接连通struct prog{int u;int v;int w;}map[80];//存储边的信息，包括起点/终点/权值bool cmp ( prog a , prog b){//排序函数，将边根据权值从小到大排return a.w<b.w;}int find(int x){//并查集的find，不解释return x==p[x]?x:p[x]=find(p[x]);}//减枝void join(int x,int y,int i){int fx=find(x),fy=find(y);if(fx!=fy){pre[fx]=fy;if(i!=-1) fee+=a[i].x;}return;}int main(){int n;while ( cin >> n , n ){int i , j ;for ( i = 0 ; i < 27 ; i ++ )p[i] = i ;//并查集初始化int k = 0 ;for ( i = 0 ; i < n - 1 ; i ++ ){//构造边的信息char str[3];int m;cin >> str >> m ;for ( j = 0 ; j < m ; j ++ ,k ++ ){char str2[3];int t;cin >> str2 >> t ;map[k].u=(str[0]-'A');map[k].v=(str2[0]-'A');map[k].w=t;}}sort ( map , map + k , cmp );//将边从小到大排序int ans=0; //所要求的答案for ( i = 0 ; i < k ; i ++ ){int x = find(map[i].u);int y = find(map[i].v);printf("%d%d\n",x,y);if( x!=y){//如果两点不在同一连通分量里，则将两点连接，并存储该边ans+=map[i].w;p[x]=y;}}cout<<ans<<endl;}return 0;}最大连续子序列-1235#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>int a[10010],maxsum,t,e,n;void ff(){int i,l,j,sum;for(i = 0; i < n; i++){sum = 0;for(l = i; l < n; l++){sum += a[l];if(sum > maxsum){maxsum = sum;t = i;e = l;// printf("%d\n",maxsum);}// printf("%d %d %d\n",i,l,sum);}}}int main(){//freopen("1235input.txt","r",stdin);int f,i;while(scanf("%d",&n)&&n){memset(a,0,sizeof(a));f =0;maxsum = -10000;for(i = 0; i < n; i++){scanf("%d",&a[i]);if(a[i] < 0)f++;}if(f == n){maxsum = 0;t = 0;e = n-1;}elseff();printf("%d %d %d\n",maxsum,a[t],a[e]);}return 0;}素数表int a[10000];//1为素数void ff(){memset(a,1,sizeof(a));for(i=2;i<=10000;i++){if(a[i]==1){printf("%d ",i);for(l=i;l<=10000;l++){if(a[l]==1)if(l%i==0)a[l]=0;}}}}================int a[10000];void ff(){memset(a,1,sizeof(a));for(int i = 2; i < 10000; i++){if(a[i] == 0){int l = i;while(i * l < 10000){a[i*l]=0;l++;} }}}BFSinclude <stdio.h>#include <string.h>int x, y, mx, my,step;char map[110][110];int move[8][2]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};int recode[110][110];int f[110][110];struct duili{int x;int y;}a[110*110];bool judge(int m, int n){if(map[m][n]== '.' &&f[m][n]==2&& m>0 &&n>0&&m<y+1&&n<x+1) return 1;elsereturn 0;}int bfs(int i,int j){int front = 0,rear = 1;step = -1;recode[i][j] = 0; f[i][j]=1;a[front].x = i;a[front].y = j;while(front < rear){int xx = a[front].x, yy = a[front].y;front++;for(i = 0; i < 8; i++ ){if(judge(xx + move[i][0],yy + move[i][1])){f[xx + move[i][0]][yy + move[i][1]] =1 ;recode[xx + move[i][0]][yy + move[i][1]] = recode[xx][yy] + 1;a[rear].x = xx + move[i][0];a[rear].y = yy + move[i][1];rear++;if (recode[xx + move[i][0]][yy + move[i][1]] > step)step = recode[xx + move[i][0]][yy + move[i][1]];}}// for(int i = y; i > 0; i--)// {// for(int j = 1; j <= x; j++)// printf("%d",f[i][j]);// printf("\n");// }// printf("\n");}return step;}int main(){scanf("%d%d%d%d",&x, &y, &mx, &my);memset(map,0,sizeof(map));memset(f,0,sizeof(f));memset(recode,0,sizeof(recode));getchar();for(int i = y; i > 0; i--){for(int j = 1; j <= x; j++){scanf("%c",&map[i][j]);if(map[i][j]=='.')f[i][j]=2;}getchar();}// for(int i = y; i > 0; i--)// {// for(int j = 1; j <= x; j++)// printf("%c",map[i][j]);// printf("\n");// }printf("%d\n",bfs(mx,my));return 0;}CD_ROOM#include <stdio.h>int maxn;int n;int arr[20];int tempmax;void ff(int cur, int sum){if (sum > maxn)return;if (cur == n){tempmax >?= sum;return;}ff(cur + 1, sum);sum += arr[cur];ff(cur + 1, sum);}int main(void){while (tempmax = 0, scanf("%d", &n) == 1) {scanf("%d", &maxn);for (int i = 0; i < n; i++)scanf("%d", arr + i);ff(0, 0);printf("%d\n", tempmax);}return 0; }Floyed(最短路径)（可以有圈）#include <iostream>#include <string>#include <map>using namespace std;const int MAXN = 31;typedef double ValueType;//数据类型void Floyed(ValueType g[][MAXN],int n)//佛洛依德算法{for(int k=0;k<n;++k)for(int i=0;i<n;++i)for(int j=0;j<n;++j)if(g[i][j]<g[i][k]*g[k][j]) g[i][j] = g[i][k]*g[k][j];//尽可能使汇率大，这里是乘法(*)}int main(){int n,m,i,Case=0;string name,sour,dest;V alueType value;map<string,int> Currency; //用map实现name和index之间的转换V alueType graph[MAXN][MAXN];while(cin>>n){if(n==0) break;Currency.clear();memset(graph,0,sizeof(graph));for(i=0;i<n;++i)//n个顶点{cin>>name;Currency[name] = i;graph[i][i] = 1;}cin>>m;//m条边for(i=0;i<m;++i)//输入边信息创建图{cin>>sour>>value>>dest;graph[Currency[sour]][Currency[dest]] = value;}Floyed(graph,n);//佛洛依德求各个节点之间的bool ok = false;for(int i=0;i<n;++i)//求自身到自身的汇率(回路) if(graph[i][i]>1){ok=true;break;}cout<<"Case "<<++Case<<": ";if(ok) cout<<"Yes\n";else cout<<"No\n";}return 0;}3个for#include<stdio.h>#define MAX 100000000int main(){int n,m,a,b,t,i,j,k,map[202][202];while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++)for(j=1;j<=n;j++)map[i][j]=MAX;while(m--){scanf("%d%d%d",&a,&b,&t);map[a][b]=t;map[b][a]=t;}for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++)if(map[i][k]+map[k][j]<map[i][j])map[i][j]=map[i][k]+map[k][j];scanf("%d%d",&a,&b);if(map[a][b]==MAX||b>n||a<1)printf("-1\n");elseprintf("%d\n",map[a][b]);}}Dijkstra算法+注释#include <iostream>#include <string.h>#include <limits.h>#include <stdio.h>//hdu1874using namespace std;int map[1002][1002];int main(){int n,m;while(~scanf("%d%d",&n,&m)){int visted[205],dis[205];memset(visted,0,sizeof(visted)); //访问数组初始化for(int i=1;i<=n;++i){dis[i]=INT_MAX; //初始化两点之间的最短路径！！for(int j=1;j<=n;++j)map[i][j]=INT_MAX; //初始化两点间距离！！map[i][i]=0;}int a,b,c;while(m--){scanf("%d%d%d",&a,&b,&c);//输入两点间最短距离if(c<map[a][b])map[a][b]=map[b][a]=c;}int begin ,end ,pos;scanf("%d%d",&begin,&end);pos=begin;visted[pos]=1;//访问初始起始点。

我的ACM算法模板

ACM模板[王克纯2020年9月21日最大子串int maxSum(int * a,int n){int sum = a[0],b = 0;for(int i=0;i<n;++i){if(b>0) b += a[i];else b = a[i];if(b > sum) sum = b;}return sum;}int Kadane(const int array[], size_t length, unsigned int& left, unsigned int& right){unsigned int i, cur_left, cur_right;int cur_max, max;cur_max = max = left = right = cur_left = cur_right = 0;for(i = 0; i < length; ++i){cur_max += array[i];if(cur_max > 0){cur_right = i;if(max < cur_max){max = cur_max;left = cur_left;right = cur_right;}}else{cur_max = 0;cur_left = cur_right = i + 1;}}return max;} 快速幂void js(int &a,int &b,int num) {b=1;while(num){if(num&1) b*=a;num>>=1;a*=a;}}矩阵乘法struct mat{int n,m;//n行m列int data[MAX][MAX];};void mul(const mat& a,const mat& b,mat& c) //c=a*b{int i,j,k;if (a.m!=b.n); //报错c.n=a.n,c.m=b.m;for (i=0;i<c.n;i++){for (j=0;j<c.m;j++){for (c.data[i][j]=k=0;k<a.m;k++) {c.data[i][j]+=a.data[i][k]*b.dat a[k][j]%m;//m为余数}c.data[i][j]%=m;}}}Bit位操作(宏定义,内联函数,stl)} #define bitwrite(a,i,n)(n)?(a)[(i)/8]|=1<<(i)%8:(a)[(i)/8]&=~(1<<(i)%8)//数组a的第i位写入n;#define bitread(a,i)((a)[(i)/8]>>((i)%8))&1//读取数组a的第i位inline void write(int i,int n){n?a[i/8]|=1<<i%8:a[i/8]&=~(1<<i% 8);}inline int read(int i){return (a[i/8]>>(i%8))&1;}#include<bitset>bitset<MAX> b;错排公式为M(n)=n！（1/2!-1/3!+…..+(-1）^n/n!)M(n)=n!-n!/1!+n!/2!-n!/3!+…+(-1）^n*n!/n!=sigma(k=2~n) (-1）^k*n!/k!Dn=[n!/e+0.5]容斥原理M(n)=n![1/0!-1/1!+1/2!-1/3!+1/4! +..+(-1）^n/n!]二分模板LL findr(LL array, LL low, LL high,LL target){while(low <= high){LL mid = (low + high)/2;if (array[mid] > target) high = mid - 1;else if (array[mid] < target) low = mid + 1;else return mid;}return -1;复用代码#include<stdio.h>#include<stdlib.h>#include<string.h>#define MAX 10void print(mat t){printf("*****************\n") ;for(int i=0;i<t.n;i++){for(int j=0;j<t.m;j++){printf("%d",t.data[i][j]);}putchar('\n');}}一些常量和函数：最大Long long __int64 INF = ~(((__int64)0x1)<<63);ceil()向上取整(math.h)floor()向下取整c字符串处理函数1）提取子串--strstr函数原型：char* strstr(char*src,char*find)函数说明：从字符串src中寻找find第一次出现的位置（不比较结束符NULL）返回值：返回指向第一次出现find位置的指针，如果没有找到则返回NULL2）接尾连接--strcat函数原型：char* strcat(char*dest,char*src)函数说明：把src所指字符串添加到dest结尾处(覆盖dest结尾处的'\0')并添加'\0'3)部分连接--strncat函数原型:char* strncat(char*dest,char*src,int n);函数说明：把src所指字符串的前n个字符添加到dest结尾处（覆盖dest结尾处的’\0’）并添加’’\0’.返回值：返回指向dest的指针。

ACM数论相关模板

求最大公约数（欧几里得算法）int gcd(int a, int b){if (b == 0) return a;return gcd(b, a % b);}求最小公倍数Lcm(a, b) = a * b / gcd(a, b);判断是否为素数#include <iostream>#include <stdio.h>#include <math.h>using namespace std;bool judgeprime(int n){bool flag =true;int m = sqrt(n);for (int i = 2; i <= m; i++){if (n % i == 0){flag = false;break;}}if (flag) printf("yes\n");else printf("no\n");}快速幂求%yx pLong long pw(long long x, long long y,long long p){if (y == 0) return 1LL;long long tmp = pw （x, y / 2， p);if (y & 1) return tmp * tmp % p * x % p;return tmp * tmp % p;}埃拉托斯特尼筛法寻找n 以内的素数 void findprime(int n){bool isprime[10005];int prime[10005], i, j, total;memset(isprime, true, sizeof(isprime));total=0;for (i = 2; i<=n; i++){if (isprime[i]){prime[total++] = i;for (j=i*i;j<=n;j+=i)isprime[j]=false;}}for (i = 0; i < total; i++)printf("%d ",prime[i]);}欧拉筛法void makePrime(){memset(isPrime,true,sizeof(isPrime));for(int i=2; i<=MAXN; i++){if(isPrime[i]) prime[total++]=i;for(int j=0; j<total && i*prime[j]<=MAX; j++){isPrime[i*prime[j]]=false;//i此时不是素数,只是拓展用if(i%prime[j]==0) break;}}}。

ACM算法--枚举方法（指数枚举，组合枚举）模板

ACM算法--枚举⽅法（指数枚举，组合枚举）模板// 递归实现指数型枚举vector<int> chosen;void calc(int x) {if (x == n + 1) {for (int i = 0; i < chosen.size(); i++)printf("%d ", chosen[i]);puts("");return;}calc(x + 1);chosen.push_back(x);calc(x + 1);chosen.pop_back();}// 递归实现组合型枚举vector<int> chosen;void calc(int x) {if (chosen.size() > m || chosen.size() + (n - x + 1) < m) return;if (x == n + 1) {for (int i = 0; i < chosen.size(); i++)printf("%d ", chosen[i]);puts("");return;}calc(x + 1);chosen.push_back(x);calc(x + 1);chosen.pop_back();}// 递归实现排列型枚举int order[20];bool chosen[20];void calc(int k) {if (k == n + 1) {for (int i = 1; i <= n; i++)printf("%d ", order[i]);puts("");return;}for (int i = 1; i <= n; i++) {if (chosen[i]) continue;order[k] = i;chosen[i] = 1;calc(k + 1);chosen[i] = 0;order[k] = 0;}}// 模拟机器实现，把组合型枚举改为⾮递归vector<int> chosen;int stack[100010], top = 0, address = 0;void call(int x, int ret_addr) { // 模拟计算机汇编指令callint old_top = top;stack[++top] = x; // 参数xstack[++top] = ret_addr; // 返回地址标号stack[++top] = old_top; // 在栈顶记录以前的top值}int ret() { // 模拟计算机汇编指令retint ret_addr = stack[top - 1];top = stack[top]; // 恢复以前的top值return ret_addr;}int main() {int n, m;cin >> n >> m;call(1, 0); // calc(1)while (top) {int x = stack[top - 2]; // 获取参数switch (address) {case 0:if (chosen.size() > m || chosen.size() + (n - x + 1) < m) {address = ret(); // returncontinue;}if (x == n + 1) {for (int i = 0; i < chosen.size(); i++)printf("%d ", chosen[i]);puts("");address = ret(); // returncontinue;}call(x + 1, 1); // 相当于calc(x + 1)，返回后会从case 1继续执⾏ address = 0;continue; // 回到while循环开头，相当于开始新的递归case 1:chosen.push_back(x);call(x + 1, 2); // 相当于calc(x + 1)，返回后会从case 2继续执⾏ address = 0;continue; // 回到while循环开头，相当于开始新的递归case 2:chosen.pop_back();address = ret(); // 相当于原calc函数结尾，执⾏return}}}。

ACM常用算法模板

专用模板目录：一、图论1．最大团2．拓扑排序3．最短路和次短路4．SAP模板5．已知各点度，问能否组成一个简单图6．KRUSKAL7. Prim算法求最小生成树8. Dijkstra9 . Bellman-ford10. SPFA11. Kosaraju 模板12. tarjan 模板二、数学1. 剩余定理2. N!中质因子P的个数3.拓展欧几里得4.三角形的各中心到顶点的距离和5.三角形外接圆半径周长6.归并排序求逆序数7. 求N！的位数8.欧拉函数9. Miller-Rabin,大整数分解，求欧拉函数10. 第一类斯特林数11.计算表达式12.约瑟夫问题13．高斯消元法14. Baby-step,giant-step n是素数.n任意15. a^b%c=a ^(b%eular(c)+eular(c)) % c16.判断第二类斯特林数的奇偶性17.求组合数C(n,r)18.进制转换19.Ronberg算法计算积分20.行列式计算21. 返回x 的二进制表示中从低到高的第i位22.高精度运算 +-*/23.超级素数筛选三、数据结构1．树状数组2.线段树求区间的最大、小值3.线段树求区间和4.单调队列5.KMP模板6. 划分树，求区间第k小数7.最大堆，最小堆模板8. RMQ模板求区间最大、最小值9.快速排序，归并排序求逆序数.10.拓展KMP四、计算几何1.凸包面积2.Pick公式求三角形内部有多少点3.多边形边上内部各多少点以及面积pick4.平面最远点对5.判断矩形是否在矩形内6.判断点是否在多边形内7.判断4个点(三维)是否共面8.凸包周长9.等周定理变形一直两端点和周长求最大面积10.平面最近点对11.单位圆最多覆盖多少点(包括边上)12.多边形费马点求点到多边形各个点的最短距离13.矩形并周长14.zoj 2500 求两球体积并一、图论1.最大团#include<iostream>#include<algorithm>using namespace std;int n,m;int cn;//当前顶点数int best;//当前最大顶点数int vis[50];//当前解int bestn[50];//最优解int map[50][50];//临界表void dfs(int i){if(i>n){for(int j=1;j<=n;j++) bestn[j]=vis[j];best=cn;return ;}int ok=1;for(int j=1;j<i;j++){if(vis[j]==1&&map[i][j]==0){ok=0;break;}}if(ok){//进入左子树vis[i]=1;cn++;dfs(i+1);cn--;}if(cn+n-i>best){//进入右子树vis[i]=0;dfs(i+1);}}int main(){while(scanf("%d%d",&n,&m)==2){memset(vis,0,sizeof(vis));memset(map,0,sizeof(map));while(m--){int p,q;scanf("%d%d",&p,&q);map[p][q]=map[q][p]=1;//无向图}cn=0;best=0;dfs(1);printf("%d\n",best);}return 0;}2.拓扑排序#include<iostream>#include<cstring>using namespace std;int map[105][105],in[105],vis[105],ans[105],n;int flag;void dfs(int step){if(flag) return ;if(step==n+1) {flag=1; printf("%d",ans[1]);for(int i=2;i<=n;i++) printf(" %d",ans[i]);printf("\n");return ;}for(int i=1;i<=n;i++){if(vis[i]==0&&in[i]==0){vis[i]=1;for(int j=1;j<=n;j++){if(map[i][j]>0){map[i][j]=-map[i][j];in[j]--;}}ans[step]=i;dfs(step+1);vis[i]=0;for(int j=1;j<=n;j++){if(map[i][j]<0){map[i][j]=-map[i][j];in[j]++;}}}}}int main(){while(scanf("%d",&n)==1){flag=0;memset(map,0,sizeof(map));memset(vis,0,sizeof(vis));memset(in,0,sizeof(in));for(int i=1;i<=n;i++){int t;while(scanf("%d",&t),t){map[i][t]=1;in[t]++;}}dfs(1);}return 0;}3．最短路和次短路#include<iostream>#include<cstdio>#include<vector>#include<cstring>using namespace std;class Node{public:int e,w;//表示终点和边权};const int inf=(1<<25);int main(){int ci;cin>>ci;while(ci--){vector<Node> G[1005];//用邻接表存边int n,m;cin>>n>>m;for(int i=1;i<=m;i++){Node q;int u;cin>>u>>q.e>>q.w;G[u].push_back(q);}int s,f;//起点和终点cin>>s>>f;//dijkstra 求最短路和次短路int flag[1005][2];int dis[1005][2],cnt[1005][2];//0表示最短路，1表示次短路memset(flag,0,sizeof(flag));for(int i=1;i<=n;i++) dis[i][0]=dis[i][1]=inf;dis[s][0]=0;cnt[s][0]=1;//初始化for(int c=0;c<2*n;c++) //找最短路和次短路，故要进行2*n次循环也可以改成while(1){int temp=inf,u=-1,k;//找s-S'集合中的最短路径,u记录点的序号，k记录是最短路或者是次短路for(int j=1;j<=n;j++){if(flag[j][0]==0&&temp>dis[j][0]) temp=dis[j][0],u=j,k=0;else if(flag[j][1]==0&&temp>dis[j][1]) temp=dis[j][1],u=j,k=1;}if(temp==inf) break;//S'集合为空或者不联通，算法结束//更新路径flag[u][k]=1;for(int l=0;l<G[u].size();l++){int d=dis[u][k]+G[u][l].w,j=G[u][l].e;//important//4种情况if(d<dis[j][0]){dis[j][1]=dis[j][0];cnt[j][1]=cnt[j][0];dis[j][0]=d;cnt[j][0]=cnt[u][k];}else if(d==dis[j][0]){cnt[j][0]+=cnt[u][k];}else if(d<dis[j][1]){dis[j][1]=d;cnt[j][1]=cnt[u][k];}else if(d==dis[j][1]){cnt[j][1]+=cnt[u][k];}}}int num=cnt[f][0];//最短路int cc=cnt[f][1];//次短路}return 0;}4．SAP模板#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int inf=(1<<31)-1;const int point_num=300;int cap[point_num][point_num],dist[point_num],gap[point_num];//初始化见main里面int s0,t0,n;//源,汇和点数int find_path(int p,int limit=0x3f3f3f3f){if(p==t0) return limit;for(int i=0;i<n;i++)if(dist[p]==dist[i]+1 && cap[p][i]>0){int t=find_path(i,min(cap[p][i],limit));if(t<0) return t;if(t>0){cap[p][i]-=t;cap[i][p]+=t;return t;}}int label=n;for(int i=0;i<n;i++) if(cap[p][i]>0) label=min(label,dist[i]+1);if(--gap[dist[p]]==0 || dist[s0]>=n ) return -1;++gap[dist[p]=label];return 0;}int sap(){//初始化s,ts0=0,t0=n-1;int t=0,maxflow=0;gap[0]=n;while((t=find_path(s0))>=0) maxflow+=t;return maxflow;}int main(){int ci;while(cin>>ci>>n){//初始化memset(cap,0,sizeof(cap));memset(dist,0,sizeof(dist));memset(gap,0,sizeof(gap));//初始化capwhile(ci--){int x,y,c;cin>>x>>y>>c;x--;y--;cap[x][y]+=c;//因题而异}int ans=sap();cout<<ans<<endl;}return 0;}5．已知各点度，问能否组成一个简单图#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int inf=(1<<30);int d[1100];bool cmp(int x,int y){return x>y;}int main(){int ci;scanf("%d",&ci);while(ci--){int n,flag=1,cnt=0;scanf("%d",&n); for(int i=0;i<n;i++){scanf("%d",&d[i]);if(d[i]>n-1||d[i]<=0) flag=0; cnt+=d[i];}if(flag==0||cnt%2){printf("no\n");continue;}sort(d,d+n,cmp);for(int l=n;l>0;l--){for(int i=1;i<l&&d[0];i++){d[0]--,d[i]--;if(d[i]<0){flag=0;break;}}if(d[0]) flag=0;if(flag==0) break;d[0]=-inf;sort(d,d+l,cmp);}if(flag) printf("yes\n");else printf("no\n");}return 0;}6.KRUSKAL#include<iostream>#include<algorithm>using namespace std;int u[15005],v[15005],w[15005],fath[15005],r[15005];int ans1[15005],ans2[15005];bool cmp(int i,int j){return w[i]<w[j];}int find(int x){return fath[x]==x?x:fath[x]=find(fath[x]);}int main(){int n,m;cin>>n>>m;for(int i=1;i<=n;i++) fath[i]=i;for(int i=1;i<=m;i++) r[i]=i;for(int i=1;i<=m;i++){cin>>u[i]>>v[i]>>w[i];}sort(r+1,r+m+1,cmp);int maxn=0,ans=0,k=0;for(int i=1;i<=m;i++){int e=r[i];int x=find(u[e]),y=find(v[e]);if(x!=y){ans+=w[e];fath[x]=y;if(w[e]>maxn) maxn=w[e];ans1[k]=u[e];ans2[k++]=v[e];}}return 0;}7.prime求最小生成树语法：prim(Graph G,int vcount,int father[]);参数：G：图，用邻接矩阵表示vcount：表示图的顶点个数father[]：用来记录每个节点的父节点返回值：null注意：常数max_vertexes 为图最大节点数常数infinity为无穷大源程序：#define infinity 1000000#define max_vertexes 5typedef int Graph[max_vertexes][max_vertexes];void prim(Graph G,int vcount,int father[]){int i,j,k;intlowcost[max_vertexes],closeset[max_vertexes],used[max_vertexes]; for (i=0;i<vcount;i++){lowcost[i]=G[0][i];closeset[i]=0;used[i]=0;father[i]=-1;}used[0]=1;for (i=1;i<vcount;i++){j=0;while (used[j]) j++;for (k=0;k<vcount;k++)if ((!used[k])&&(lowcost[k]<lowcost[j])) j=k;father[j]=closeset[j];used[j]=1;for (k=0;k<vcount;k++)if (!used[k]&&(G[j][k]<lowcost[k])){ lowcost[k]=G[j][k];closeset[k]=j; }}}8.Dijkstra语法：result=Dijkstra(Graph G,int n,int s,int t, int path[]); 参数：G：图，用邻接矩阵表示n：图的顶点个数s：开始节点t：目标节点path[]：用于返回由开始节点到目标节点的路径返回值：最短路径长度注意：输入的图的权必须非负顶点标号从0 开始用如下方法打印路径：i=t;while (i!=s){printf("%d<--",i+1);i=path[i];}printf("%d\n",s+1);源程序：int Dijkstra(Graph G,int n,int s,int t, int path[]){int i,j,w,minc,d[max_vertexes],mark[max_vertexes];for (i=0;i<n;i++) mark[i]=0;for (i=0;i<n;i++){ d[i]=G[s][i];path[i]=s; }mark[s]=1;path[s]=0;d[s]=0;for (i=1;i<n;i++){minc=infinity;w=0;for (j=0;j<n;j++)if ((mark[j]==0)&&(minc>=d[j])) {minc=d[j];w=j;}mark[w]=1;for (j=0;j<n;j++)if((mark[j]==0)&&(G[w][j]!=infinity)&&(d[j]>d[w]+G[w][j])){ d[j]=d[w]+G[w][j];path[j]=w; }}return d[t];}9.Bellman-ford语法：result=Bellman_ford(Graph G,int n,int s,int t,int path[],int success);参数：G：图，用邻接矩阵表示n：图的顶点个数s：开始节点t：目标节点path[]：用于返回由开始节点到目标节点的路径success：函数是否执行成功返回值：最短路径长度注意：输入的图的权可以为负，如果存在一个从源点可达的权为负的回路则success=0顶点标号从0 开始用如下方法打印路径：i=t;while (i!=s){printf("%d<--",i+1);i=path[i];}printf("%d\n",s+1);源程序：int Bellman_ford(Graph G,int n,int s,int t,int path[],int success){int i,j,k,d[max_vertexes];for (i=0;i<n;i++) {d[i]=infinity;path[i]=0;}d[s]=0;for (k=1;k<n;k++)for (i=0;i<n;i++)for (j=0;j<n;j++)if (d[j]>d[i]+G[i][j]){d[j]=d[i]+G[i][j];path[j]=i;}success=0;for (i=0;i<n;i++)for (j=0;j<n;j++)if (d[j]>d[i]+G[i][j]) return 0;success=1;return d[t];}10. SPFA#include<iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;const __int64 maxn=1001000;const __int64 inf=1000100000;struct edge//邻接表{__int64 t,w;//s->t=w;__int64 next;//数组模拟指针};__int64 p[maxn],pf[maxn];//邻接表头节点edge G[maxn],Gf[maxn];//邻接表__int64 V,E;//点数[1-n] 边数__int64 dis[maxn];__int64 que[maxn],fro,rear;//模拟队列__int64 vis[maxn];__int64 inque[maxn];//入队次数bool spfa(__int64 s0){fro=rear=0;for(__int64 i=1;i<=V;i++) dis[i]=inf;dis[s0]=0;memset(vis,0,sizeof(vis));memset(inque,0,sizeof(inque));que[rear++]=s0;vis[s0]=1;inque[s0]++;while(fro!=rear){__int64 u=que[fro];fro++;if(fro==maxn) fro=0;vis[u]=0;for(__int64 i=p[u];i!=-1;i=G[i].next){__int64 s=u,t=G[i].t,w=G[i].w;if(dis[t]>dis[s]+w){dis[t]=dis[s]+w;if(vis[t]==0){que[rear++]=t,vis[t]=1;inque[t]++;if(inque[t]>V) return false;if(rear==maxn) rear=0;}}}}return true;}int main(){__int64 ci;scanf("%I64d",&ci);while(ci--){scanf("%I64d%I64d",&V,&E);memset(p,-1,sizeof(p));memset(pf,-1,sizeof(pf)); for(__int64 i=0;i<E;i++){__int64 u,v,w;scanf("%I64d%I64d%I64d",&u,&v,&w);G[i].t=v;G[i].w=w;G[i].next=p[u];p[u]=i;Gf[i].t=u;Gf[i].w=w;Gf[i].next=pf[v];pf[v]=i;}__int64 ans=0;spfa(1);//求第一个点到其他点的最短距离和for(__int64 i=1;i<=V;i++) ans+=dis[i];//反方向再来一次spfa 求其他点到第一个点的最短距离和 for(__int64 i=1;i<=V;i++) p[i]=pf[i];for(__int64 i=0;i<E;i++) G[i]=Gf[i];spfa(1);for(__int64 i=1;i<=V;i++) ans+=dis[i];printf("%I64d\n",ans);}return 0;}11.Kosaraju模板#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100000;struct edge{int t,w;//u->t=w;int next;};int V,E;//点数(从1开始)，边数int p[maxn],pf[maxn];//邻接表原图,逆图edge G[maxn],Gf[maxn];//邻接表原图,逆图int l,lf;void init(){memset(p,-1,sizeof(p));memset(pf,-1,sizeof(pf));l=lf=0;}void addedge(int u,int t,int w,int l){G[l].w=w;G[l].t=t;G[l].next=p[u];p[u]=l;}void addedgef(int u,int t,int w,int lf){Gf[l].w=w;Gf[l].t=t;Gf[l].next=pf[u];pf[u]=l;}///Kosaraju算法,返回为强连通分量个数bool flag[maxn]; //访问标志数组int belg[maxn]; //存储强连通分量,其中belg[i]表示顶点i属于第belg[i]个强连通分量int numb[maxn]; //结束时间(出栈顺序)标记,其中numb[i]表示离开时间为i的顶点//用于第一次深搜,求得numb[1..n]的值void VisitOne(int cur, int &sig){flag[cur] = true;for (int i=p[cur];i!=-1;i=G[i].next){if (!flag[G[i].t]){VisitOne(G[i].t,sig);}}numb[++sig] = cur;}//用于第二次深搜,求得belg[1..n]的值void VisitTwo(int cur, int sig){flag[cur] = true;belg[cur] = sig;for (int i=pf[cur];i!=-1;i=Gf[i].next){if (!flag[Gf[i].t]){VisitTwo(Gf[i].t,sig);}}//Kosaraju算法,返回为强连通分量个数int Kosaraju_StronglyConnectedComponent(){int i, sig;//第一次深搜memset(flag,0,sizeof(flag));for ( sig=0,i=1; i<=V; ++i ){if ( false==flag[i] ){VisitOne(i,sig);}}//第二次深搜memset(flag,0,sizeof(flag));for ( sig=0,i=V; i>0; --i ){if ( false==flag[numb[i]] ){VisitTwo(numb[i],++sig);}}return sig;}int main(){while(scanf("%d",&V)==1){init();for(int i=1;i<=V;i++){int u=i,t,w=1;while(scanf("%d",&t)==1&&t){E++;addedge(u,t,w,l++);addedgef(t,u,w,lf++);}}int ans=Kosaraju_StronglyConnectedComponent(); printf("%d\n",ans);}return 0;12.tarjan模板//自己模板#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100000;int V,E;//点数(1) 边数struct edge//邻接表{int t,w;//u->t=w;int next;};int p[maxn];//表头节点edge G[maxn];int l;void init(){memset(p,-1,sizeof(p));l=0;}//添加边void addedge(int u,int t,int w,int l)//u->t=w;{G[l].w=w;G[l].t=t;G[l].next=p[u];p[u]=l;}//tarjan算法求有向图强联通分量int dfn[maxn],lowc[maxn];//dfn[u]节点u搜索的次序编号,lowc[u]u或者u的子树能够追溯到的栈中的最早的节点int belg[maxn];//第i个节点属于belg[i]个强连通分量int stck[maxn],stop;//stck栈int instck[maxn];//第i个节点是否在栈中int scnt;//强联通分量int index;void dfs(int i){dfn[i]=lowc[i]=++index;instck[i]=1;//节点i入栈stck[++stop]=i;for(int j=p[i];j!=-1;j=G[j].next){int t=G[j].t;//更新lowc数组if(!dfn[t])//t没有遍历过{dfs(t);if(lowc[i]>lowc[t]) lowc[i]=lowc[t];}//t是i的祖先节点else if(instck[t]&&lowc[i]>dfn[t]) lowc[i]=dfn[t];}//是强连通分量的根节点if(dfn[i]==lowc[i]){scnt++;int t;do{t=stck[stop--];instck[t]=0;belg[t]=scnt;}while(t!=i);}}int tarjan(){stop=scnt=index=0;memset(dfn,0,sizeof(dfn));memset(instck,0,sizeof(instck));for(int i=1;i<=V;i++){if(!dfn[i]) dfs(i);}return scnt;}int main(){while(scanf("%d",&V)==1){init();for(int i=1;i<=V;i++){int x;while(scanf("%d",&x)==1&&x){E++;addedge(i,x,1,l++);}}int ans=tarjan();printf("%d\n",ans);}return 0;}//吉大模板邻接表版#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100000;int V,E;//点数(1) 边数struct edge//邻接表{int t,w;//u->t=w;int next;};int p[maxn];//表头节点edge G[maxn];int l;void init(){memset(p,-1,sizeof(p));l=0;}//添加边void addedge(int u,int t,int w,int l)//u->t=w;{G[l].w=w;G[l].t=t;G[l].next=p[u];p[u]=l;}//tarjan算法求有向图强联通分量int dfn[maxn],lowc[maxn];//dfn[u]节点u搜索的次序编号,lowc[u]u或者u的子树能够追溯到的栈中的最早的节点int stck[maxn],stop;//stck栈int pre[maxn];//int scnt;//强联通分量int cnt;//void dfs(int v)//1-V{int t,minc=lowc[v]=pre[v]=cnt++;stck[stop++]=v;for(int i=p[v];i!=-1;i=G[i].next){int pv=G[i].t;if(pre[pv]==-1) dfs(pv);if(lowc[pv]<minc) minc=lowc[pv]; }if(minc<lowc[v]){lowc[v]=minc;return ;}do{dfn[t=stck[--stop]]=scnt;lowc[t]=V;}while(t!=v);++scnt;}int tarjan(){stop=cnt=scnt=0;memset(pre,-1,sizeof(pre));for(int i=1;i<=V;i++){if(pre[i]==-1) dfs(i);}return scnt;}int main(){while(scanf("%d",&V)==1){init();for(int i=1;i<=V;i++){int x;while(scanf("%d",&x)==1&&x){E++;addedge(i,x,1,l++);}}int ans=tarjan();printf("%d\n",ans);}return 0;}二、数学1．剩余定理int mod(int c[],int b[],int n){int all_multy=1,sum=0;int i,j,x[5];for(i=0;i<n;i++)all_multy*=c[i];for(i=0;i<n;i++)x[i]=all_multy/c[i];for(i=0;i<n;i++){j=1;while((x[i]*j)%c[i]!=1)j++;x[i]*=j;}for(i=0;i<n;i++)sum+=(b[i]*x[i]);return sum%all_multy;}2．N!中质因子P的个数//对于任意质数p，n！中有（n/p+n/p^2+n/p^3+...)个质因子p。

-【精品资料】ACM大赛必备_常用函数整理_ACM模板(整理版)

目录一、数学问题 (4)1.精度计算——大数阶乘 (4)2.精度计算——乘法（大数乘小数） (4)3.精度计算——乘法（大数乘大数） (5)4.精度计算——加法 (6)5.精度计算——减法 (7)6.任意进制转换 (8)7.最大公约数、最小公倍数 (9)8.组合序列 (10)9.快速傅立叶变换（FFT） (10)10.Ronberg 算法计算积分 (12)11.行列式计算 (14)12.求排列组合数 (15)13.求某一天星期几 (15)14.卡特兰(Catalan) 数列原理 (16)15.杨辉三角 (16)16.全排列 (17)17.匈牙利算法----最大匹配问题 (18)18.最佳匹配KM 算法 (20)二、字符串处理 (22)1.字符串替换 (22)2.字符串查找 (23)3.字符串截取 (24)4.LCS-最大公共子串长度 (24)5.LCS-最大公共子串长度 (25)6.数字转换为字符 (26)三、计算几何 (27)1.叉乘法求任意多边形面积 (27)2.求三角形面积 (27)3.两矢量间角度 (28)4.两点距离（2D、3D） (28)5.射向法判断点是否在多边形内部 (29)6.判断点是否在线段上 (30)7.判断两线段是否相交 (31)8.判断线段与直线是否相交 (32)9.点到线段最短距离 (32)10.求两直线的交点 (33)11.判断一个封闭图形是凹集还是凸集 (34)12.Graham 扫描法寻找凸包 (35)13.求两条线段的交点 (36)四、数论 (37)1.x 的二进制长度 (37)2.返回x 的二进制表示中从低到高的第i 位 (38)3.模取幂运算 (38)4.求解模线性方程 (39)5.求解模线性方程组(中国余数定理) (39)6.筛法素数产生器 (40)7.判断一个数是否素数 (41)8.求距阵最大和 (42)8.求一个数每一位相加之和 (43)10.质因数分解 (43)11.高斯消元法解线性方程组 (44)五、图论 (45)1.Prim 算法求最小生成树................................................. 45 2.Dijkstra 算法求单源最短路径.. (46)3.Bellman-ford 算法求单源最短路径 (47)4.Floyd-Warshall 算法求每对节点间最短路径 (48)5.解欧拉图 (49)六、排序/查找 (50)1.快速排序 (50)2.希尔排序 (51)3.选择法排序 (52)4.二分查找 (52)七、数据结构 (53)1.顺序队列 (53)2.顺序栈 (56)3.链表 (59)4.链栈 (63)5.二叉树 (66)八、高精度运算专题 (68)1.专题函数说明 (68)2.高精度数比较 (69)3.高精度数加法 (69)4.高精度数减法 (70)5.高精度乘10 (71)6.高精度乘单精度 (71)7.高精度乘高精度 (72)8.高精度除单精度 (72)9.高精度除高精度 (73)九、标准模板库的使用 (74)1.计算求和 (74)2.求数组中的最大值 (76)3. sort 和qsort (76)十、其他 (78)1.运行时间计算 (78)DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD一、数学问题1.精度计算——大数阶乘语法：int result=factorial(int n);参数：n：n 的阶乘返回值：阶乘结果的位数注意：本程序直接输出n!的结果，需要返回结果请保留long a[] 需要math.h源程序：int factorial(int n){long a[10000];int i,j,l,c,m=0,w;a[0]=1;for(i=1;i<=n;i++){c=0;for(j=0;j<=m;j++){a[j]=a[j]*i+c;c=a[j]/10000;a[j]=a[j]%10000;}if(c>0) {m++;a[m]=c;}}w=m*4+log10(a[m])+1;printf("\n%ld",a[m]);for(i=m-1;i>=0;i--) printf("%4.4ld",a[i]);return w;}我也可以做到..5 / 782.精度计算——乘法（大数乘小数）语法：mult(char c[],char t[],int m);参数：c[]：被乘数，用字符串表示，位数不限t[]：结果，用字符串表示m：乘数，限定10 以内返回值：null注意：需要string.h源程序：void mult(char c[],char t[],int m){int i,l,k,flag,add=0;char s[100];l=strlen(c);for (i=0;i<l;i++)s[l-i-1]=c[i]-'0';for (i=0;i<l;i++){k=s[i]*m+add;if (k>=10) {s[i]=k%10;add=k/10;flag=1;} else{s[i]=k;flag=0;add=0;}}if (flag) {l=i+1;s[i]=add;} else l=i;for (i=0;i<l;i++)t[l-1-i]=s[i]+'0'; t[l]='\0';}3.精度计算——乘法（大数乘大数）语法：mult(char a[],char b[],char s[]);参数：a[]：被乘数，用字符串表示，位数不限b[]：乘数，用字符串表示，位数不限t[]：结果，用字符串表示返回值：null注意：空间复杂度为o(n^2)需要string.h源程序：void mult(char a[],char b[],char s[]){我也可以做到..6 / 78int i,j,k=0,alen,blen,sum=0,res[65][65]={0},flag=0; char result[65];alen=strlen(a);blen=strlen(b);for (i=0;i<alen;i++)for (j=0;j<blen;j++) res[i][j]=(a[i]-'0')*(b[j]-'0');for (i=alen-1;i>=0;i--){for (j=blen-1;j>=0;j--) sum=sum+res[i+blen-j-1][j]; result[k]=sum%10;k=k+1;sum=sum/10;}for (i=blen-2;i>=0;i--){for (j=0;j<=i;j++) sum=sum+res[i-j][j];result[k]=sum%10;k=k+1;sum=sum/10;}if (sum!=0) {result[k]=sum;k=k+1;}for (i=0;i<k;i++) result[i]+='0';for (i=k-1;i>=0;i--) s[i]=result[k-1-i];s[k]='\0';while(1){if (strlen(s)!=strlen(a)&&s[0]=='0')strcpy(s,s+1);elsebreak;}}4.精度计算——加法语法：add(char a[],char b[],char s[]);参数：a[]：被加数，用字符串表示，位数不限b[]：加数，用字符串表示，位数不限s[]：结果，用字符串表示返回值：null注意：空间复杂度为o(n^2)我也可以做到..7 / 78需要string.hDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD源程序：void add(char a[],char b[],char back[]){int i,j,k,up,x,y,z,l;char *c;if (strlen(a)>strlen(b)) l=strlen(a)+2; else l=strlen(b)+2; c=(char *) malloc(l*sizeof(char));i=strlen(a)-1;j=strlen(b)-1;k=0;up=0;while(i>=0||j>=0){if(i<0) x='0'; else x=a[i];if(j<0) y='0'; else y=b[j];z=x-'0'+y-'0';if(up) z+=1;if(z>9) {up=1;z%=10;} else up=0;c[k++]=z+'0';i--;j--;}if(up) c[k++]='1';i=0;c[k]='\0';for(k-=1;k>=0;k--)back[i++]=c[k];back[i]='\0';}5.精度计算——减法语法：sub(char s1[],char s2[],char t[]);参数：s1[]：被减数，用字符串表示，位数不限s2[]：减数，用字符串表示，位数不限t[]：结果，用字符串表示返回值：null注意：默认s1>=s2，程序未处理负数情况需要string.h源程序：void sub(char s1[],char s2[],char t[])我也可以做到..8 / 78{int i,l2,l1,k;l2=strlen(s2);l1=strlen(s1);t[l1]='\0';l1--;for (i=l2-1;i>=0;i--,l1--){if (s1[l1]-s2[i]>=0)t[l1]=s1[l1]-s2[i]+'0';else{t[l1]=10+s1[l1]-s2[i]+'0';s1[l1-1]=s1[l1-1]-1;}}k=l1;while(s1[k]<0) {s1[k]+=10;s1[k-1]-=1;k--;}while(l1>=0) {t[l1]=s1[l1];l1--;}loop:if (t[0]=='0') {l1=strlen(s1);for (i=0;i<l1-1;i++) t[i]=t[i+1];t[l1-1]='\0';goto loop;}if (strlen(t)==0) {t[0]='0';t[1]='\0';}}6.任意进制转换语法：conversion(char s1[],char s2[],char t[]);参数：s[]：转换前的数字s2[]：转换后的数字d1：原进制数d2：需要转换到的进制数返回值：null注意：高于9 的位数用大写'A'～'Z'表示，2～16 位进制通过验证源程序：void conversion(char s[],char s2[],long d1,long d2){我也可以做到..9 / 78long i,j,t,num;char c;num=0;for (i=0;s[i]!='\0';i++){if (s[i]<='9'&&s[i]>='0') t=s[i]-'0'; else t=s[i]-'A'+10;num=num*d1+t;}i=0;while(1){t=num%d2;if (t<=9) s2[i]=t+'0'; else s2[i]=t+'A'-10;num/=d2;if (num==0) break;i++;}for (j=0;j<i/2;j++){c=s2[j];s2[j]=s[i-j];s2[i-j]=c;}s2[i+1]='\0';}7.最大公约数、最小公倍数语法：resulet=hcf(int a,int b)、result=lcd(int a,int b)参数：a：int a，求最大公约数或最小公倍数b：int b，求最大公约数或最小公倍数返回值：返回最大公约数（hcf）或最小公倍数（lcd）注意：lcd 需要连同hcf 使用源程序：int hcf(int a,int b){int r=0;while(b!=0){r=a%b;a=b;DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDb=r;}return(a);我也可以做到..10 / 78}lcd(int u,int v,int h){return(u*v/h);}8.组合序列语法：m_of_n(int m, int n1, int m1, int* a, int head)参数：m：组合数C 的上参数n1：组合数C 的下参数m1：组合数C 的上参数，递归之用*a：1～n 的整数序列数组head：头指针返回值：null注意：*a 需要自行产生初始调用时，m=m1、head=0调用例子：求C(m,n)序列：m_of_n(m,n,m,a,0);源程序：void m_of_n(int m, int n1, int m1, int* a, int head){int i,t;if(m1<0 || m1>n1) return;if(m1==n1){return;}m_of_n(m,n1-1,m1,a,head); // 递归调用t=a[head];a[head]=a[n1-1+head];a[n1-1+head]=t;m_of_n(m,n1-1,m1-1,a,head+1); // 再次递归调用t=a[head];a[head]=a[n1-1+head];a[n1-1+head]=t;}9.快速傅立叶变换（FFT）语法：kkfft(double pr[],double pi[],int n,int k,double fr[],double fi[],intl,int il);参数：我也可以做到..11 / 78pr[n]：输入的实部pi[n]：数入的虚部n，k：满足n=2^kfr[n]：输出的实部fi[n]：输出的虚部l：逻辑开关，0 FFT，1 ifFTil：逻辑开关，0 输出按实部/虚部；1 输出按模/幅角返回值：null注意：需要math.h源程序：void kkfft(pr,pi,n,k,fr,fi,l,il)int n,k,l,il;double pr[],pi[],fr[],fi[];{int it,m,is,i,j,nv,l0; double p,q,s,vr,vi,poddr,poddi;for (it=0; it<=n-1; it++){m=it; is=0;for (i=0; i<=k-1; i++){j=m/2; is=2*is+(m-2*j); m=j;}fr[it]=pr[is]; fi[it]=pi[is];}pr[0]=1.0; pi[0]=0.0;p=6.283185306/(1.0*n);pr[1]=cos(p); pi[1]=-sin(p);if (l!=0) pi[1]=-pi[1];for (i=2; i<=n-1; i++){p=pr[i-1]*pr[1];q=pi[i-1]*pi[1];s=(pr[i-1]+pi[i-1])*(pr[1]+pi[1]);pr[i]=p-q; pi[i]=s-p-q;}for (it=0; it<=n-2; it=it+2){vr=fr[it]; vi=fi[it];fr[it]=vr+fr[it+1]; fi[it]=vi+fi[it+1];fr[it+1]=vr-fr[it+1]; fi[it+1]=vi-fi[it+1]; }m=n/2; nv=2;for (l0=k-2; l0>=0; l0--){我也可以做到..12 / 78m=m/2; nv=2*nv;for (it=0; it<=(m-1)*nv; it=it+nv)for (j=0; j<=(nv/2)-1; j++){p=pr[m*j]*fr[it+j+nv/2];q=pi[m*j]*fi[it+j+nv/2];s=pr[m*j]+pi[m*j];s=s*(fr[it+j+nv/2]+fi[it+j+nv/2]); poddr=p-q; poddi=s-p-q;fr[it+j+nv/2]=fr[it+j]-poddr;fi[it+j+nv/2]=fi[it+j]-poddi;fr[it+j]=fr[it+j]+poddr;fi[it+j]=fi[it+j]+poddi;}}if (l!=0)for (i=0; i<=n-1; i++){fr[i]=fr[i]/(1.0*n);fi[i]=fi[i]/(1.0*n);}if (il!=0)for (i=0; i<=n-1; i++){pr[i]=sqrt(fr[i]*fr[i]+fi[i]*fi[i]);if (fabs(fr[i])<0.000001*fabs(fi[i])) {if ((fi[i]*fr[i])>0) pi[i]=90.0;else pi[i]=-90.0;}DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDelsepi[i]=atan(fi[i]/fr[i])*360.0/6.283185306;}return;}10.Ronberg 算法计算积分语法：result=integral(double a,double b);参数：a：积分上限b：积分下限我也可以做到..13 / 78function f：积分函数返回值：f 在（a,b）之间的积分值注意：function f(x)需要自行修改，程序中用的是sina(x)/x 需要math.h默认精度要求是1e-5源程序：double f(double x){return sin(x)/x; //在这里插入被积函数}double integral(double a,double b){double h=b-a;double t1=(1+f(b))*h/2.0;int k=1;double r1,r2,s1,s2,c1,c2,t2;loop:double s=0.0;double x=a+h/2.0;while(x<b){s+=f(x);x+=h;}t2=(t1+h*s)/2.0;s2=t2+(t2-t1)/3.0;if(k==1){k++;h/=2.0;t1=t2;s1=s2;goto loop;}c2=s2+(s2-s1)/15.0;if(k==2){c1=c2;k++;h/=2.0;t1=t2;s1=s2;goto loop;}r2=c2+(c2-c1)/63.0;if(k==3){r1=r2; c1=c2;k++;h/=2.0;t1=t2;s1=s2;我也可以做到..14 / 78goto loop;}while(fabs(1-r1/r2)>1e-5){ r1=r2;c1=c2;k++;h/=2.0;t1=t2;s1=s2;goto loop;}return r2;}11.行列式计算语法：result=js(int s[][],int n)参数：s[][]：行列式存储数组n：行列式维数，递归用返回值：行列式值注意：函数中常数N 为行列式维度，需自行定义源程序：int js(s,n)int s[][N],n;{int z,j,k,r,total=0;int b[N][N];/*b[N][N]用于存放，在矩阵s[N][N]中元素s[0]的余子式*/if(n>2){for(z=0;z<n;z++){for(j=0;j<n-1;j++)for(k=0;k<n-1;k++)if(k>=z) b[j][k]=s[j+1][k+1]; elseb[j][k]=s[j+1][k];if(z%2==0) r=s[0][z]*js(b,n-1); /*递归调用*/else r=(-1)*s[0][z]*js(b,n-1);total=total+r;}}else if(n==2)total=s[0][0]*s[1][1]-s[0][1]*s[1][0];return total;我也可以做到..15 / 78}12.求排列组合数语法：result=P(long n,long m); / result=long C(long n,long m);参数：m：排列组合的上系数n：排列组合的下系数返回值：排列组合数注意：符合数学规则：m<＝n源程序：long P(long n,long m){long p=1;while(m!=0){p*=n;n--;m--;}return p;}long C(long n,long m){long i,c=1;DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDi=m;while(i!=0){c*=n;n--;i--;}while(m!=0){c/=m;m--;}return c;}13.求某一天星期几语法：result=weekday(int N,int M,int d)参数：N,M,d：年月日，例如：2003,11,4返回值：0：星期天，1 星期一……注意：需要math.h适用于1582 年10 月15 日之后, 因为罗马教皇格里高利十三世在这一天启用新历法.源程序：我也可以做到..16 / 78int weekday(int N,int M,int d){int m,n,c,y,w;m=(M-2)%12;if (M>=3) n=N;else n=N-1;c=n/100;y=n%100;w=(int)(d+floor(13*m/5)+y+floor(y/4)+floor(c/4)-2*c)%7;while(w<0) w+=7;return w;}14.卡特兰(Catalan) 数列原理令h(1)＝1，catalan 数满足递归式：h(n)= h(1)*h(n-1) + h(2)*h(n-2) + ... + h(n-1)h(1) (其中n>=2)该递推关系的解为：h(n)=c(2n-2,n-1)/n (n=1,2,3,...)1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440,9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420,24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, …1.括号化问题。

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算法模板
Wenen
2017-9-29始
一万年太久，只争朝夕
高精度加法——Java版本
//输入两个数
//输出两数相加结果
import java.util.*;
import java.math.*;
publicclass Main{
publicstaticvoid main(String[] args)
{
Scanner cin=new Scanner(System.in);//检测键盘是否有输入
BigInteger ans=BigInteger.valueOf(0);//答案为大数
while(cin.hasNext())//如果有下一行
{
BigInteger a=cin.nextBigInteger();//等于下一行
BigInteger b=cin.nextBigInteger();
ans=a.add(b);//加法
System.out.println(ans);//输出答案
}
}
}
闰年：同时满足以下条件：1、年份能被4整除；2、年份若是100的整数倍的
话,需被400整除,否则是平年.举例如下：1900年能被4整除,但是因为其是100的整数倍,却不能被400整除,所以是平年；而2000年就是闰年
闰年函数：
bool is_leap_year(int y)
{
if(y%100==0)
{
if(y%400==0)return true;
else return false;
}
else if(y%4==0)
{
return true;
}
else
{
return false;
}
}
最大质因数函数：
int largest_prime_factor(int n)
{
if (n<1) return -1;
//判断边界条件
if (n==1)return 1;
while (n >1)
{
for(int i=2;i<=n;i++)
{
if (n==i)//到达n了，就没有继续的必要了，已经最大
return n;
if(n%i==0)//
{
n = n/i;
break;
}
}
}
}
九余数定理
一个数对九取余后的结果称为九余数。

一个数的各位数字相加后得到的<10的数字称为这个数的九余数（如果相加结果大于9，则继续各位相加）
我么们知道，不管是什么数，数九余数一定是0~9其中的一个，因为不会有正数各位之和为0。

证明：
假设，数d的九余数为d%9( 暂时不取0，整除时取9）
当d < 10时，1~9这9个数肯定成立;
当d >= 10时，d的九余数为d%9 = (d-1)%9+1，即d的前一个数的九余数加1.
得证.
N^N的九余数代码：
for(int i=1;i<=n;i++)
ans=ans*n%9;
if(ans==0)cout<<9<<endl;
else cout<<ans<<endl;
质因数分解
#include <stdio.h>
int main()
{
int data, i = 2;
scanf("%d", &data);
while(data > 1)
{
if(data % i == 0)
{
printf("%d ", i); data /= i;
}
else i++;
}
}
数论规律：
尾数为0,1,5,6的不管是多少次方尾数依然不变，而尾数为4和9的每2次循环，
2,3,7,8为每4次循环。

循环结果如下：
0，1，5，6：位数永远是0，1，5，6
2：6，2，4，8循环
3：1，3，9，7循环
4：6，4循环
7：1，7，9，3循环
8：6，8，4，2循环
9：1，9循环
以上全为从0次起。