1 /* 2 Problem:cf187 div2 D 3 Type:string 4 Time:2013.06.09 5 //思路，直接统计b字符串每个字符开始，经过一个a字符串的结果 6 所谓结果，a的一个循环包含多少个当前字符后的字符 7 保存完后对a统计即可 8 好吧，够绕口的，直接看代码比较清楚 9 */ 10 #include <iostream> 11 #include <cstdio> 12 #include <cstdlib> 13 #include <cstring> 14 #include <cmath> 15 #include <algorithm> 16 #include <string> 17 #include <stack> 18 #include <set> 19 #include <queue> 20 #include <map> 21 #include <fstream> 22 #include <vector> 23 #include <list> 24 #define MaxPoint 20000 25 #define MaxLine 50 26 #define MaxNum 1999997 27 #define Inf 0xfffff 28 #define M0(a) memset(a, 0, sizeof(a)) 29 using namespace std; 30 int n, m; 31 char a[200], b[200]; 32 int next[200], cnt[200]; 33 void solve(){ 34 scanf("%s%s", &a, &b); 35 int la = strlen(a), lb = strlen(b), cur; 36 37 for (int i = 0; i < lb; ++i){ 38 cur = i; 39 for (int j = 0; j < la; ++j) //从cur开始在一个周期的a的作用下能到达哪里，并统计当前答案，记住next数组 40 if (a[j] == b[cur]){ 41 ++cur; 42 if (cur == lb){ 43 ++cnt[i]; 44 cur = 0; 45 } 46 } 47 next[i] = cur; 48 } 49 long long ans = cur = 0; 50 for (int i = 1; i <= n; ++i){ //直接计算每个a数组的答案 51 ans += cnt[cur]; 52 cur = next[cur]; 53 } 54 printf("%I64d\n", ans / m); 55 } 56 57 int main(){ 58 freopen("a.in","r", stdin); 59 freopen("a.out","w", stdout); 60 scanf("%d%d", &n, &m); 61 solve(); 62 fclose(stdin); fclose(stdout); 63 }