hdu 5793 A Boring Question（2016第六场多校）

A Boring Question

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 487    Accepted Submission(s):
271

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

 

题意：根据题目中规定的结算方式，给定n,m的值，求结果。

 

打表找规律，比赛的时候找到了还是没写出来。。因为有除法还有取模，所以要用到逆元，第一次使用逆元，找了一个模板。

 

附上代码：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #define ll long long
 #define mod 1000000007
 using namespace std;

 ll mat(ll a,ll b)///a^b，结果对mod取膜，b的值很大的时候
 {
     ll c=;
     if(b==) return a%mod; ///当b为1时，只剩下最后一个a
     else if(b&)  ///b为奇数
         return mat(a,b-)*a%mod; ///把单独的a拿出来
     else ///b为偶数
         return mat(a*a%mod,b/)%mod; ///直接相乘，系数除以2
 }

 ll extend_gcd(ll a,ll b,ll &x,ll &y)
 {
     if(a==&&b==) return -;//无最大公约数
     if(b==){x=;y=;return a;}
     ll d=extend_gcd(b,a%b,y,x);
     y-=a/b*x;
     return d;
 }
 //*********求逆元素*******************
 //ax = 1(mod n)
 ll mod_reverse(ll a,ll n)
 {
     ll x,y;
     ll d=extend_gcd(a,n,x,y);
     if(d==) return (x%n+n)%n;
     else return -;
 }

 ll c(ll n,ll m)
 {
     ll i,j,t1,t2,ans;
     t1=((mat(m,n+)-)%mod+mod)%mod;
     t2=(m-)%mod;
     return  t1*mod_reverse(t2,mod)%mod;
 }

 int main()
 {
     int T,n,m;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         ll ans=c(n,m);
         printf("%I64d\n",ans);
     }
     return ;
 }