hdu-5793 A Boring Question(二项式定理)

题目链接：

A Boring Question

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)


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

[align=left]Sample Input[/align]
[align=left] [/align]

2
1 2
2 3

[align=left]Sample Output[/align]
[align=left] [/align]

3
13

题意：

就是求这个式子的值是多少;

思路：

∑(km,km-1)(km-1,km-2)...(k2,k1)=∑(km,km-1)...(k3,k2)(∑(k2,k1){0<=k1<=k2})=∑(km,km-1)...∑(k3,k2)*2k2
∑(k3,k2)*2k2 =(1+2)k3;二项式定理,以后也是这样,最后得到的结果为(mn+1-1)/(m-1);

AC代码：

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=2e3+14;
const double eps=1e-12;

LL pow_mod(LL x,LL y)
{
LL s=1,base=x;
while(y)
{
if(y&1)s=s*base%mod;
base=base*base%mod;
y>>=1;
}
return s;
}

int main()
{
int t;
read(t);
while(t--)
{
LL n,m;
read(n);read(m);
cout<<(pow_mod(m,n+1)-1+mod)%mod*pow_mod(m-1,mod-2)%mod<<"\n";
}
return 0;
}