BZOJ 3884 上帝与集合的正确使用方法

#扩展欧拉定理 ![](https://img2018.cnblogs.com/blog/1564177/201908/1564177-20190818101207267-684413441.png)

\(gcd(a,m)=1\)时就是欧拉定理

后面两种情况对\(gcd\)无要求,注意\(c\)\(\phi(m)\)的关系

\(Description\)

\(2^{2^{2^{2^{\cdots}}}}(mod p)\)
简化为求\(x\equiv 2^x (mod  p))\)

\(Solution\)

因为\(2^{2^{2^{2^{\cdots}}}} > \phi (p)\)
\(\phi(p)\)在不断缩小,所以用第三个公式即可
\(ans(p)\)表示对\(p\)取模的答案,显然\(ans(p)=2^{ans(\phi(p))+\phi(p)}\)
\(\phi(p)\)在不断缩小,直到为\(1\)时,则任何数\(mod 1=0\),所以到达递归边界,返回\(0\)即可

\(Code\)

#include <cstdio>
#define int long long

typedef long long ll;
const int N = 1e7+71;

int p_(ll x, int y, int z) {
    int ret = 1;
    for (; y; (x *= x) %= z, y >>= 1) if (y & 1) (ret *= x) %= z;
    return ret;
}
signed phi[N];
ll solve(int p) {
    if (p == 1) return 0;
    return p_(2, solve(phi[p]) + phi[p], p);
}
int t, p;

signed main() {
    phi[1] = 1;
    for (signed i = 2; i < N; i++) if (!phi[i]) {
        for (signed j = i; j < N; j += i) {
            if (!phi[j]) phi[j] = j;
            phi[j] = phi[j] / i * (i-1);
        }
    }
    for (scanf("%lld", &t); t--; ) {
        scanf("%lld", &p);
        printf("%lld\n", solve(p));
    }
}
posted @ 2019-08-18 10:20  __Liuz  阅读(116)  评论(0编辑  收藏  举报