hdu5793——A Boring Question（快速幂+逆元）

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

打表推推推，最后发现是等比数列求和

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <map>#include <set>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 100010#define mod 1000000007using namespace std;long long PowerMod(long long a, long long b, long long c){    long long ans = 1;    a = a % c;    while(b>0)    {        if(b % 2 == 1)            ans = (ans * a) % c;        b = b/2;        a = (a * a) % c;    }    return ans;}long long extend_gcd(long long a,long long b,long long &x,long long &y)//ax+by=1返回a,b的gcd，同时求的一组满足题目的最小正整数解{    long long ans,t;    if(b==0)    {        x=1;        y=0;        return a;    }    ans=extend_gcd(b,a%b,x,y);    t=x;    x=y;    y=t-(a/b)*y;    return ans;}int main(){    long long t,n,m,ans,x,y;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d",&n,&m);        long long a=PowerMod(m,n+1,mod)-1,b=m-1;        extend_gcd(b,mod,x,y);        while(x<0)            x+=mod;        ans=a*x%mod;        printf("%I64d\n",ans);    }    return 0;}