Exponial （欧拉定理+指数循环定理+欧拉函数+快速幂）

题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=2021

Description

Everybody loves big numbers (if you do not, you might want to stop reading at this point). There are many ways of constructing really big numbers known to humankind, for instance:

• Exponentiation: 422016=42⋅42⋅...⋅42����������������2016 times422016=42⋅42⋅...⋅42⏟2016 times.
• Factorials: 2016!=2016 ⋅ 2015 ⋅ ... ⋅ 2 ⋅ 1.

In this problem we look at their lesser-known love-child the exponial, which is an operation defined for all positive integers n​ as
exponial(n)=n^{(n − 1)(n − 2)⋯21}
For example, exponial(1)=1 and exponial(5)=5⁴³²¹ ≈ 6.206 ⋅ 10¹⁸³²³⁰ which is already pretty big. Note that exponentiation is right-associative: a^bc = a^(bc).

Since the exponials are really big, they can be a bit unwieldy to work with. Therefore we would like you to write a program which computes exponial(n) mod m (the remainder of exponial(n) when dividing by m).

Input

There will be several test cases. For the each case, the input consists of two integers n (1 ≤ n ≤ 10⁹) and m (1 ≤ m ≤ 10⁹).

Output

Output a single integer, the value of exponial(n) mod m.

Sample Input

2 42
5 123456789
94 265

Sample Output

2
16317634
39

思路：本题是一道经典的指数循环定理简记e(n)=exponial(n)e(n)=exponial(n)，利用欧拉定理进行降幂即可，不过要注意会爆int。指数循环公式为指数循环公式为A^B = A^(B %  φ(C) +  φ(C)) % C,其中 φ(C)为1~C-1中与C互质的数的个数。

代码如下：

#include <cstdio>
#include <cstring>

typedef long long ll;
int n, m;

ll euler(int n) {
    ll ans = n;
    for(int i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            ans = ans / i * (i - 1);
            while(n % i == 0) n /= i;
        }
    }
    if(n > 1) ans = ans / n * (n - 1);
    return ans;
}

ll ModPow(int x, int p, ll mod) {
    ll rec = 1;
    while(p > 0) {
        if(p & 1) rec = (ll)rec * x % mod;
        x = (ll)x * x % mod;
        p >>= 1;
    }
    return rec;
}

ll slove(int n, ll m) {
    if(m == 1) return 0;
    if(n == 1) return 1 % m;
    if(n == 2) return 2 % m;
    if(n == 3) return 9 % m;
    if(n == 4) return (1 << 18) % m;
    return (ll)ModPow(n, euler(m), m) * ModPow(n, slove(n - 1, euler(m)), m) % m;
}

int main() {
    while(~scanf("%d%d", &n, &m)) {
        printf("%lld
",slove(n, m));
    }
    return 0;
}

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