HDU 5793 A Boring Question (找规律 : 快速幂+乘法逆元)

A Boring Question

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 865    Accepted Submission(s): 534

Problem Description

There are an equation.
∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=?
We define that (kj+1kj)=kj+1!kj!(kj+1−kj)! . And (kj+1kj)=0 while kj+1<kj.
You have to get the answer for each n and m that given to you.
For example,if n=1,m=3,
When k1=0,k2=0,k3=0,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=0,k3=0,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=0,(k2k1)(k3k2)=0;
Whenk1=0,k2=0,k3=1,(k2k1)(k3k2)=1;
Whenk1=0,k2=1,k3=1,(k2k1)(k3k2)=1;
Whenk1=1,k2=0,k3=1,(k2k1)(k3k2)=0;
Whenk1=1,k2=1,k3=1,(k2k1)(k3k2)=1.
So the answer is 4.

 

Input

The first line of the input contains the only integer T,(1≤T≤10000)
Then T lines follow,the i-th line contains two integers n,m,(0≤n≤109,2≤m≤109)

 

Output

For each n and m,output the answer in a single line.

 

Sample Input

2
1 2
2 3

 

Sample Output

3
13

 

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6

Recommend

wange2014

题解：

找规律...
f(0,2)=1; f(1,2)=3; f(2,2)=7; f(3,2)=15
f(0,3)=1; f(1,3)=4; f(2,3)=13; 
f(0,4)=1; f(1,4)=5; f(2,4)=21;
f(0,5)=1; f(1,5)=6; f(2,5)=31;

f(n,m)=f(n-1,m)*m+1

所以 f(n,m)=(m^(n+1)-1)/(m-1)

//#include <iostream>

#include<bits/stdc++.h>

using namespace std;

const long long mod=;

long long ans,n,m;

int T;

long long poww(long long a,long long b)

{

    long long ans=;

    while(b)

    {

        if (b%==) ans=(ans*a)%mod;

        a=(a*a)%mod;

        b/=;

    }

    return ans;

}

int main()

{

   scanf("%d",&T);

   for(;T>;T--)

   {

       scanf("%lld%lld",&n,&m);

       long long ans=poww(m,n+);

       ans=(ans+mod-)%mod;

       ans=(ans*poww(m-,mod-))%mod;

       printf("%lld\n",ans);

   }

   return ;

}