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 = ; i*i <= c; i++) {
         if(c%i == ){
             ans -= ans/i;
             while(c%i == ) c /= i;
         }
     }
     if(c > ) 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 = ;
     while(b){
         if(b&){
             ans = ans*a;
             if(ans >= mod) ans = ans%mod + mod;　　//**
         }
         a *= a;
         if(a >= mod) a = a%mod + mod;　　//**
         b >>= ;
     }
     return ans;
 }
 ll solve(ll n, ll m)
 {
     ll p = phi(m);
     if(n == ) return ;
     ll index = solve(n/, p);

     return quick_mod(n%, index, m);
 }
 int main()
 {
     ll n, m, T;
     cin >> T;
     while(T--){
         scanf("%I64d%I64d", &n, &m);
         printf("%I64d\n", solve(n,m)%m);
     }
     return ;
 }