hdu 2837 坑题。

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1414    Accepted Submission(s): 291

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

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

 /**
 a ^ b % c= a ^ ( b % phi ( c )  +  phi ( c ) ) % c      条件:（b>=c）  phi(n)为n的欧拉函数值
 我想说(⊙o⊙)…，完全搞不懂标程
 **/
 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;
 typedef __int64 LL;

 LL Euler(LL n)
 {
     LL temp =n,i;
     for(i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             while(n%i==)
                 n=n/i;
             temp=temp/i*(i-);
         }
     }
     if(n!=) temp=temp/n*(n-);
     return temp;
 }
 LL pow_mod(LL a,LL b,LL p){
     LL ans=;
     while(b)
     {
         if(b&) ans=(ans*a)%p;
         b=b>>;
         a=(a*a)%p;
     }
     return ans;
 }
 LL fuck(LL a,LL n,LL m)
 {
     LL i,ans=;
     for(i=;i<=n;i++)
     {
         ans=ans*a;
         if(ans>=m) return ans;
     }
     return ans;
 }
 LL dfs(LL n,LL m){
    LL ans,p,nima;
    if(n==) return ;
    if(n<) return n;
    p = Euler(m);
    ans=dfs(n/,p);

    nima = fuck(n%,ans,m);
    if(nima>=m){
         ans=pow_mod(n%,ans%p+p,m);
         if(ans==) return m;
    }
    else{
        ans=pow_mod(n%,ans,m);
    }
    return ans;
 }
 int main()
 {
     int T;
     LL n,m;
     scanf("%d",&T);
     while(T--){
         scanf("%I64d%I64d",&n,&m);
         printf("%I64d\n",dfs(n,m)%m);
     }
     return ;
 }