uva 11127(暴力)

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;
}