exercise3

第三次作业1、两名参与者交替从一堆石子中取出若干数目,其个数由参与者自已决定.但是要求参与者每次至少取出一个,至多取出一半,然后另一名参与者继续.拿到最后一个石子的参与者将输掉该游戏.输入：输入开始时石子的个数输出：如果先取者输，则输出“lose”，否则输出“win”#include<iostream>#include<cmath>using namespace std;int main(){int number;cin>>number;double i=1;double k=pow(2.0,i)-1;while(k<number){++i;k=pow(2.0,i)-1;}if(k==number)cout<<"lose";if(k>number)cout<<"win";return 0;}2、Given some Chinese Coins (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum valuethat you cannot pay with given coins.InputInput contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.OutputOutput the minimum positive value that one cannot pay with given coins, one line for one case.#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAX = 1000;const int type[3] = {1,2,5};bool ans[MAX]={0},tmp[MAX]={0};int cnt[3];void Search(){int i,j,k;for(i=0;i<=cnt[0];++i){ans[i]=true;}for(i=1;i<3;++i){for(j=0;j<MAX;++j){for(k=0;k<=cnt[i] && j+k*type[i]<=MAX;++k){if(tmp[j+k*type[i]] || ans[j])tmp[j+k*type[i]]=true;}}for(j=1;j<=MAX;++j){ans[j]=tmp[j];tmp[j]=false;}}}int main(){int res=0;scanf("%d %d %d",&cnt[0],&cnt[1],&cnt[2]);if(cnt[0]==0 && cnt[1]==0 && cnt[2]==0){printf("0");return 0;}Search();for(int i=0;i<=MAX;++i){if(!ans[i]){res=i;break;}}printf("%d",res);return 0;}3、写出完整的二分查找程序，要求：（1）先将给定的数据利用快速排序进行排序（采用复杂度为nlg(n)的算法）。

（2）利用二分查找法判断给定的数据是否在数据集中。

#include<iostream>#include<cstdio>using namespace std;void quicksort(int a[],int left,int right){if(left>=right)return;else{int first=left;int last=right;int key=a[first];while(first<last){while(first<last && a[last]>=key)--last;a[first]=a[last];while(first<last && a[first]<key)++first;a[last]=a[first];}a[first]=key;quicksort(a,left,first-1);quicksort(a,first+1,right);}}int search(int num,int a[],int size){int left=0;int right=size-1;int mid=(left+right)/2;while(left<=right){if(a[mid]==num)return mid;else if(a[mid]>num)right=mid-1;elseleft=mid+1;mid=(left+right)/2;}return 0;}int main(){int s[]={2,1,5,87,43,23,9};quicksort(s,0,6);int num=23;if(search(num,s,7))printf("给定的数据在数据集中!");elseprintf("给定的数据不在数据集中!");return 0;}4、Problem DescriptionYou are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.InputThe first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.OutputThere should be one line per test case containing the length of the largest string found.Sample Input23ABCDBCDFFBRCD2roseorchidSample Output22#include<iostream>#include<cstring>#include<cstdio>using namespace std;char minstr[100];//最小字符串char str[100][100];//储存所有字符串char temp[100];//将要比较的字符串char strvel[100];//逆序的字符串int s[100];int result[100];int t=0;void Vel(char p[]){int Len=strlen(p);int i,k=Len;for(i=0;i<k;++i){strvel[i]=p[--Len];}strvel[k]='\0';}int Find(int num){int minLength=strlen(minstr);int i,j;int k=minLength;while(k){for(int i=0;i+k<=minLength;++i){strncpy(temp,minstr+i,k);temp[k]='\0';Vel(temp);for(j=0;j<num;++j){if(!strstr(str[j],temp) && !strstr(str[j],strvel))break;if(j==num-1){result[t++]=k;return k;}}temp[0]='\0';}--k;}result[t++]=k;return k;}int main(){int kind,number;int i;scanf("%d",&kind);while(kind--){int minLen=100;scanf("%d",&number);for(i=0;i<number;++i){scanf("%s",&str[i]);if(minLen>strlen(str[i])){minLen=strlen(str[i]);strcpy(minstr,str[i]);}}Find(number);}int j;for(j=0;j<t;++j){printf("%d\n",result[j]);}return 0;}5、用母函数求方程x1+x2+...+xk=m的所有正整数解的个数。

输入：两个整数k和m。

输出：方程解的个数。

要求构造相应的母函数，并实现两个多项式相乘的函数。