HDU_5793_ABoringQuestion（二项式展开）

A Boring Question

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 544    Accepted Submission(s): 313

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;
When k1=0,k2=1,k3=0,(k2k1)(k3k2)=0;
When k1=1,k2=0,k3=0,(k2k1)(k3k2)=0;
When k1=1,k2=1,k3=0,(k2k1)(k3k2)=0;
When k1=0,k2=0,k3=1,(k2k1)(k3k2)=1;
When k1=0,k2=1,k3=1,(k2k1)(k3k2)=1;
When k1=1,k2=0,k3=1,(k2k1)(k3k2)=0;
When k1=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

题意……

解题思路

其实比赛的的时候哪有什么思路

本来想生推公式但是发现

除了找到非递降序列才有值这一有用规律就再不能看出什么了

也忘记了交换计算次序

从而变成二项式展开的形式

那就直接打表找规律呗……

ABORINGPROBLEM

正经化简如官方题解所示

1、非递降序列有值

2、交换计算次序，把乘法对应的项提到自己做变元的那个累加号后

3、后面的就是二项式展开反着用往前推即可

    例如最后一个式子那里sigma(2^k2*C(k3,k2))=(1+2)^k3

4、注意最后一次化简到的式子不能用二项式定理了，是一个等比数列求和

5、化成公式，费马小处理逆元，问题完美得解

#include 
#include 
#include 
using namespace std;

const int MO=1e9+7;
typedef long long LL;
LL qm(LL a,LL b)
{
    LL ans=0;
    while(b)
    {
        if(b&1)
            ans=(ans+a)%MO;
        a=(a+a)%MO;
        b>>=1;
    }
    return ans;
}

LL qp(LL a,LL n)
{
    LL ans=1;
    while(n)
    {
        if(n&1)
            ans=(ans*a)%MO;
        a=(a*a)%MO;
        n>>=1;
    }
    return ans;
}

int main()
{
    int t;
    int m,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf("%I64d\n",(qp(m,n+1)-1+MO)%MO*qp(m-1,MO-2)%MO);
    }

    return 0;
}