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 <stdio.h>

#include <string.h>

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;

}