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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
For each n and m,output the answer in a single line.

[align=left]Sample Input[/align]

2
1 2
2 3

[align=left]Sample Output[/align]

3
13

[align=left]Author[/align]
UESTC

[align=left]Source[/align]
2016 Multi-University Training Contest 6
[align=left]Recommend[/align]
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=1000000007;
long long ans,n,m;
int T;
long long poww(long long a,long long b)
{
long long ans=1;
while(b)
{
if (b%2==1) ans=(ans*a)%mod;
a=(a*a)%mod;
b/=2;
}
return ans;
}

int main()
{
scanf("%d",&T);
for(;T>0;T--)
{
scanf("%lld%lld",&n,&m);
long long ans=poww(m,n+1);
ans=(ans+mod-1)%mod;
ans=(ans*poww(m-1,mod-2))%mod;
printf("%lld\n",ans);
}
return 0;
}