acm入门基础题解一

Problem A: 数字三角形#include<stdio.h>#include<stdlib.h>constintmaxn=110;int a[maxn][maxn],b[maxn][maxn],n; voiddata_set(){for(int i=1;i<=n;i++){for(int j=1;j<=i;j++){scanf("%d",&a[i][j]);}}}void solve(){for(int j=1;j<=n;j++)b[n][j]=a[n][j];for(int i=n-1;i>=1;i--)for(int j=1;j<=i;j++){if(b[i+1][j+1]>b[i+1][j])b[i][j]=b[i+1][j+1]+a[i][j];elseb[i][j]=b[i+1][j]+a[i][j];}printf("%d\n",b[1][1]);}int main(){while(scanf("%d",&n)!=EOF&&n!=0){data_set();solve();}return 0;}Problem B: 去北京看奥运#include<stdio.h>#include<string.h>constintmaxn=110;constintinf=200000000;int a[maxn],b[maxn][maxn],dp[maxn][maxn],n; voiddata_set(){for(int j=0;j<maxn;j++)for(int k=0;k<maxn;k++)dp[k][j]=inf;dp[0][1]=0;ints,e,l;scanf("%d",&n);a[0]=1;a[n+1]=1;for(int i=1;i<=n;i++) scanf("%d",&a[i]);for(int i=0;i<=n;i++){for(int j=1;j<maxn;j++)for(int k=1;k<maxn;k++)b[k][j]=inf;while(1){scanf("%d",&s);if(s==0) break;scanf("%d%d",&e,&l);b[s][e]=l;}for(int j=1;j<=a[i+1];j++){for(int k=1;k<=a[i];k++){if(dp[i+1][j]>dp[i][k]+b[k][j]&&b[k][j]!=inf) dp[i+1][j]=dp[i][k]+b[k][j];}}}printf("%d\n",dp[n+1][1]);}int main(void){int t;scanf("%d",&t);while(t--)data_set();}Problem C: 计算直线的交点数#include <iostream>#include <set>using namespace std;int main(){intn,tmp;set<int> s[21];s[0].insert(0);s[1].insert(0);for(int i=2;i<21;i++){for(int j=0;j<i;j++){set<int>::iterator it;for(it=s[j].begin();it!=s[j].end();it++){tmp=*it+(i-j)*j;s[i].insert(tmp);}}}while(cin>>n){set<int>::iterator it;for(it=s[n].begin();it!=s[n].end();it++){if(it!=s[n].begin()) cout<<" ";cout<<*it;}cout<<endl;}}Problem D: 免费馅饼#include<stdio.h>#include<string.h>int a[100050][11];int main(){int i, j, n, p, q, x, m;while(scanf("%d", &n)!=EOF){if(n==0) break;m=0;memset(a, 0, sizeof(a));for(i=1;i<=n;i++){scanf("%d%d", &q, &p);a[p][q]++;if(p>m) m=p;}for(i=m;i>=0;i--){for(j=0;j<11;j++){int x=a[i+1][j];if(j>0&&a[i+1][j-1]>x) x=a[i+1][j-1]; if(j<10&&a[i+1][j+1]>x) x=a[i+1][j+1]; a[i][j]+=x;}}printf("%d\n", a[0][5]);}return 0;}Problem E: 地道战#include<stdio.h>#include<string.h>constintmaxn=110;constintinf=200000000;intn,m,a[maxn][maxn],r[maxn][maxn],d[maxn][maxn]; int min(inta,int b){return a<b?a:b;}voiddata_set(){for(int i=1;i<=n;i++)for(int j=1;j<m;j++)scanf("%d",&r[i][j]);for(int i=1;i<=m;i++)for(int j=1;j<n;j++)scanf("%d",&d[j][i]);for(int i=0;i<=n;i++)for(int j=0;j<=m;j++)a[i][j]=inf;}void solve(){a[1][1]=0;for(int i=2;i<=m;i++) a[1][i]=a[1][i-1]+r[1][i-1];for(int i=2;i<=n;i++) a[i][1]=a[i-1][1]+d[i-1][1];for(int i=2;i<=n;i++){for(int j=2;j<=m;j++){a[i][j]=min(a[i][j-1]+r[i][j-1],a[i-1][j]+d[i-1][j]);}}printf("%d\n",a[n][m]);}int main(){while(scanf("%d%d",&n,&m)!=EOF){data_set();solve();}}Problem F: 最长不下降序列#include<stdio.h>#include<stdlib.h>#include<iostream>using namespace std;constintmaxn=2000;inti,j,n,t,x,lmax,a[maxn],b[maxn];int main(){int max;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)cin>>a[i];b[n]=1;lmax=0;for(i=n-1;i>=1;i--){max=0;for(j=i+1;j<=n;j++)if(a[i]<=a[j]&&b[j]>max)max=b[j];b[i]=max+1;if(b[i]>lmax) lmax=b[i];}cout<<lmax<<endl;}return 0;}Problem A: Substrings#include<iostream>#include<algorithm>#include<string>using namespace std;bool comp(string str1,string str2){ return str1.length()<str2.length(); }stringstr[112];int main(){int t;cin>>t;while(t--){int n;cin>>n;int i=0;for(i=0;i<n;i++)cin>>str[i];sort(str,str+n,comp);string::iterator p1;string::iterator p2;string::iterator p0; intans=0,max=0;for(p1=str[0].begin();p1<str[0].end();p1++) for(p2=str[0].end()-1;p2>=p1;p2--){string ss1=str[0].substr(p1-str[0].begin(),p2-p1+1); string ss2;for(string::iterator it=p2;it>=p1;it--){ss2+=*it;}ans=ss1.length();int flag=1;for(i=1;i<n;i++){int x=str[i].find(ss1);int y=str[i].find(ss2);if(x==-1&&y==-1){flag=0;break;}}if(flag==1&&max<=ans)max=ans;}cout<<max<<endl;}return 0;}ProblemB:CallingExtraterrestrialIntelligcn#include<cstdio>#include<iostream>#include<cstring>#define N 100005using namespace std;boolno_prim[N];intm,a,b,p,q;voidprimetable(){inti,j;for(i=3;i<317;i+=2){if(no_prim[i]==false){inttmp=i<<1;for(j=i*i;j<N;j+=tmp)no_prim[j]=true;}}}int main(){inti,j;primetable();while(scanf("%d%d%d",&m,&a,&b)&&m&&a&&b){p=q=0;double limit=a*1.0/b;for(i=2;i<=m;i++){if(!no_prim[i]&&i%2!=0||i==2){for(j=2;j<=i&&i*j<=m;j++){if(!no_prim[j]&&j%2!=0||j==2){if(j*1.0/i>=limit&&i*j>p*q) //i和j的顺序{p=j;q=i;}}}}}printf("%d %d\n",p,q);}return 0;}Problem C: 深入浅出学算法028-桥本分数式#include <stdio.h>int main(){inti,k,g,s;int m1,m2,m3,a[10];a[1]=1;i=1;g=1;s=0;while(1){g=1;for(k=i-1;k>0;k--)if(a[k]==a[i]) {g=0; break;}if(i==9 && g==1 && a[1]<a[4]){m1=a[2]*10+a[3];m2=a[5]*10+a[6];m3=a[8]*10+a[9];if(a[1]*m2*m3+a[4]*m1*m3==a[7]*m1*m2){s++;}}if(i<9 &&g==1){i++; a[i]=1; continue;}while(a[i]==9 && i>1) i--;if(a[i]==9 && i==1) break;else a[i]++;}printf("%d\n",s);}Problem D: 深入浅出学算法029-素数和环#include<stdio.h>#include<math.h>int main(){intt,i,j,n,k,s,a[2000],b[1000]; while(~scanf("%d",&n)){for(k=1;k<=2*n;k++) b[k]=0; for(k=3;k<=2*n;k+=2){for(t=0,j=3;j<=sqrt(k);j+=2)if(k%j==0){t=1;break;}if(t==0) b[k]=1;}a[1]=1;s=0;i=2;a[i]=2;while(1){t=1;for(j=1;j<i;j++)if(a[j]==a[i] || b[a[i]+a[i-1]]!=1){t=0;break;}if(t && i==n && b[a[n]+1]==1){s++;}if(t && i<n) {i++;a[i]=2;continue;} while(a[i]==n) i--;if(i>1) a[i]++;else break;}printf("%d\n",s);}return 0;}。