摘要：n*n中选8个-->每行皇后选一列，用c[x]来存储列编号，枚举量为8！=40320void search(int cur){ if(cur == n) tot ++; else for(int i = 0; i < n; i ++) { int ok =... 阅读全文

摘要：1.增量构造法view codevoid print_subset(int n, int* a, int cur){ for(int i = 0; i < cur; i++) printf("%d",a[i]); //递归一次，打印一次 prin... 阅读全文

摘要：//枚举#include int a[100];int n;void premutation(int n,int* a,int cur){ if(cur == n) { for(int i = 0; i const int M... 阅读全文

摘要：#include #include char s[100005],t[100005];int main(){ int yes, n = 0; while(scanf("%s%s",s,t)==2) { yes = 1; n = 0; for... 阅读全文

摘要：关键是判断循环的长度，若某一步的余数已出现过，则循环确定#include #include int s[10000], y[10000],sign[10000];int main(){ int a, b, sz, yu, k , cnt; while(scanf("%d%d",&a,&b... 阅读全文

摘要：本题只有四个字符，用常量数组存储它们#include #include char a[100][1500];int b[10];int main(){ int cnt, m, n, hamming= 0,maxum = 0; char c[10] ="ACGT"; scanf("%... 阅读全文

摘要：我的思路是网格的四周全部置为'*'，设置sign数组记录起始格#include #include char a[20][20];int sign[20][20];int n = 1;int main(){ int r, c, cnt, yes, first = 1; while(scan... 阅读全文

摘要：Puzzle A children's puzzle that was popular 30 years ago consisted of a 5x5 frame which contained 24 small squares of equal size. A unique letter of t 阅读全文

摘要：UVA - 455Periodic StringsTime Limit:3000MSMemory Limit:Unknown64bit IO Format:%lld & %lluSubmitStatusDescriptionPeriodic StringsA character string is ... 阅读全文