HDU - 2837 Calculation （欧拉降幂，a ^ (b ^ c)以及更高）

Calculation

Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10) for all n bigger than zero. Please calculate f(n)%m. (2 ≤ n , m ≤ 10^9, x^y means the y th power of x).

Input

The first line contains a single positive integer T. which is the number of test cases. T lines follows.Each case consists of one line containing two positive integers n and m.

Output

One integer indicating the value of f(n)%m.

Sample Input

2
24 20
25 20

Sample Output

16
5

思路：由欧拉降幂公式有，

故有，假设都 小于对应的 欧拉函数值.

所以在快速幂的时候不是直接取模，而是   % mod + mod ....  f() 函数求得是次幂，所以是在快速幂的时候做操作

Code：

#include<bits/stdc++.h>
#define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl
#define pii pair<int,int>
#define clr(a,b) memset((a),b,sizeof(a))
#define rep(i,a,b) for(int i = a;i < b;i ++)
#define pb push_back
#define MP make_pair
#define LL long long
#define ull unsigned LL
#define ls i << 1
#define rs (i << 1) + 1
#define INT(t) int t; scanf("%d",&t)

using namespace std;

LL Euler(LL n){
    LL ans = n;
    for(LL 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 pow(LL a,LL b,LL mod){
    LL ans = 1;
    while(b){
        if(b & 1){
            ans *= a;
            if(ans >= mod)
                ans = ans % mod + mod;
        }
        b >>= 1;
        a *= a;
        if(a >= mod)
            a = a % mod + mod;
    }
    return ans;
}

LL f(LL n,LL phim){
    if(n == 0){
        return 1LL;
    }
    LL tmp = f(n / 10,Euler(phim));
    return pow(n % 10,tmp,phim);
}

int main() {
    int t; scanf("%d",&t);
    while(t --){
        LL n,m; scanf("%lld%lld",&n,&m);
        printf("%lld\n",f(n,m) % m);
    }
    return 0;
}