hdu2837 Calculation a^b%p=a^(b%phi(p)+phi(p))%p

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 873    Accepted Submission(s): 192

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

224 2025 20

 

Sample Output

165

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

Recommend

gaojie

 

a^b%p=a^(b%phi(p)+phi(p))%p   b>=phi(p)

#include<iostream>#include<cstdlib>#include<stdio.h>#define ll __int64using namespace std;ll powermod(ll a,ll b,ll p){    ll res=1;    while(b)    {        if(b&1)res=res*a%p;        b>>=1;        a=a*a%p;    }    return res;}ll get_phi(int m){    ll res=1;    for(int i=2;i*i<=m;i++)    {        if(m%i==0)        {            m/=i;            res*=(i-1);            while(m%i==0){m/=i;res*=i;}        }    }    if(m!=1)    res*=(m-1);    return res;}ll check(int a,int b,int p){    ll res=1;    for(int i=1;i<=b;i++)    {        res*=a;        if(res>=p) return res;    }    return res;}

ll dfs(int n,int m){    ll phi=get_phi(m);    if(n<10)    return n;    ll x=dfs(n/10,phi);    ll yy=check(n%10,x,m);    if(yy>=m)    {        ll res=powermod(n%10,x+phi,m);        if(res==0)        res+=m;        return res;    }    else    return yy;}

int main(){    int n,m,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        printf("%I64d\n",dfs(n,m)%m);    }    return 0;}