HDU 2837 Calculation

Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2772    Accepted Submission(s): 673

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

 

Recommend

gaojie

题意：给一个公式f(n)=f(n%10)^f(n/10)，f(0)=1，求f(n)

f(n)只能递归计算，但是x^y中y的值可能过大，这里要用到一个公式x^y%m=x^(y%phi(m)+phi(m))%m，可参考：http://blog.csdn.net/doyouseeman/article/details/51863382，当x与m互质时就是一种特殊情况，x^phi(m)=1，这个公式只能在y>phi(m)的时候使用

#pragma comment(linker,"/STACK：1024000000,1024000000”)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
ll n,m;
int t;

ll ouler(ll x)
{
ll ans=x;
for(ll i=2;sqr(i)<=x;i++)
if(!(x%i))
{
ans=ans/i*(i-1);
while(!(x%i))x/=i;
}
if(x>1)ans=ans/x*(x-1);
return ans;
}

ll quick(ll n,ll p,ll m)
{
ll ans=1;
while(p)
{
if(p&1)
{
ans=ans*n;
if(ans>m)
ans=ans%m+m;
}
n=n*n;
if(n>m)n=n%m+m;
p>>=1;
}
return ans;
}

ll solve(ll n,ll m)
{
if(n<10)return n;
ll p=solve(n/10,ouler(m));
return quick(n%10,p,m);
}

int main()
{
sf("%d",&t);
while(t--)
sf("%I64d%I64d",&n,&m),pf("%I64d\n",solve(n,m)%m);
return 0;
}