hdu 2837 Calculation 指数循环节套路题

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2272    Accepted Submission(s): 536

Problem Description

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

 

题意：求解$f(n) = {(n\%10)}^{f(n/10)}\%m $，其中$(2 <= n, m <= 10^9)$;

思路：一看到指数模除，并且没有说明是否互素，直接上指数循环节，化简指数==套路...

注：

1. 使用指数循环节都是递归形式，所以欧拉函数也需要是递归形式；

2. 在指数循环节中的快速幂时，需要在ans >= mod时，模完之后还要加上mod;

#include <iostream>
#include<cstdio>
using namespace std;
#define ll long long
ll phi(ll c)
{
    ll ans = c;
    for(int i = 2; i*i <= c; i++) {
        if(c%i == 0){
            ans -= ans/i;
            while(c%i == 0) c /= i;
        }
    }
    if(c > 1) ans -= ans/c;
    return ans;
}
ll quick_mod(ll a, ll b, ll mod)
{
    if(a >= mod) a = a%mod + mod;   // 并不是直接%mod 
    ll ans = 1;
    while(b){
        if(b&1){
            ans = ans*a;
            if(ans >= mod) ans = ans%mod + mod;　　//**
        }
        a *= a;
        if(a >= mod) a = a%mod + mod;　　//**
        b >>= 1;
    }
    return ans;
}
ll solve(ll n, ll m)
{
    ll p = phi(m);
    if(n == 0) return 1;
    ll index = solve(n/10, p);
    
    return quick_mod(n%10, index, m);
}
int main()
{
    ll n, m, T;
    cin >> T;
    while(T--){
        scanf("%I64d%I64d", &n, &m);
        printf("%I64d
", solve(n,m)%m);
    }
    return 0;
}