题意：给出一个字符串，包含0、1、*，当中×是能够替换成0或者1的，假设字符串的某个子串S有SSS这种连续反复3次出现，不是Triple-free串，问给出的字符串能够形成多少个非Triple-free串。
题解：由于串长度最多31，所以能够暴力枚举每一位，边枚举边推断。#include#include const int N = 35;char str[N], str2[N];int n;long long res;bool judge(int cur) { for (int i = 1; i * 3 <= (cur + 1); i++) { int e = cur - i * 3, cnt2 = 0; for (int j = cur; j > cur - i; j--) { int cnt = 0; for (int k = j; k > e; k -= i) if (str2[j] != str2[k]) break; else cnt++; if (cnt == 3) cnt2++; else break; } if (cnt2 == i) return false; } return true;}void dfs(int cur) { if (cur == n) { res++; return; } if (cur == 0 || cur == 1) { if (str[cur] == '0' || str[cur] == '1') { str2[cur] = str[cur]; dfs(cur + 1); } else { str2[cur] = '0'; dfs(cur + 1); str2[cur] = '1'; dfs(cur + 1); } return; } str2[cur] = '0'; if (judge(cur)) { if (str[cur] == '0' || str[cur] == '*') dfs(cur + 1); } str2[cur] = '1'; if (judge(cur)) { if (str[cur] == '1' || str[cur] == '*') dfs(cur + 1); }}int main() { int cas = 1; while (scanf("%d", &n) == 1 && n) { scanf("%s", str); res = 0; dfs(0); printf("Case %d: %lld\n", cas++, res); } return 0;}