GYM 101889J(枚举、环上gcd)

答案只有n - 1种暴举即可,对于每种,gcd是一那踩雷稳了,否则看雷的分布有没有把模余占满。

const int maxn = 1e5 + 5;
int n, ans;
char str[maxn];
vector<int> pos;
bool vis[maxn], yes[maxn];

bool ok(int t) {
    if (vis[t]) return yes[t];
    vis[t] = true;
    set<int> s;
    for (int i : pos) {
        s.insert(i % t);
        if (s.size() == t)  return yes[t] = false;
    }
    return yes[t] = true;
}

int __gcd(int a, int b) {
    return b ? __gcd(b, a % b) : a;
}

int main() {
    scanf("%s", str);
    n = strlen(str);

    for (int i = 0; str[i]; ++i) {
        if (str[i] == 'P') {
            pos.push_back(i);
        }
    }

    if (pos.size() == 0)    ans = n - 1;
    else {
        for (int i = 1; i < n; ++i) {
            int tmp = __gcd(i, n);
            if (tmp > 1) {
                if (ok(tmp))  ans++;
            }
        }  
    }

    writeln(ans);
    return 0;
}
posted @ 2019-04-08 22:19  AlphaWA  阅读(164)  评论(0编辑  收藏  举报