多校6 A Boring Question 5793

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最新推荐文章于 2020-01-26 23:23:47 发布

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分类专栏：快速幂求逆元组合数

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本文探讨了一种特定的组合数学问题，即计算给定n和m时特定等式的解的数量，并提供了一个高效的算法实现方案。该算法利用了快速幂运算来解决大规模数据输入的情况。

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A Boring Question

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 635 Accepted Submission(s): 375

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

(1≤T≤10000)
Then

T lines follow,the i-th line contains two integers

(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

代码：

#include <algorithm>
#include <cstdio>
#include <vector>
#include <iostream>

using namespace std;

const int mod = 1000000007;
int n,m;

 long long quickmod(long long a,long long b)
{
    long long ans = 1;
    while(b)//用一个循环从右到左便利b的所有二进制位
    {
        if(b&1)//判断此时b[i]的二进制位是否为1
        {
            ans = (ans*a)%mod;//乘到结果上，这里a是a^(2^i)%m
            b--;//把该为变0
        }
        b/=2;
        a = a*a%mod;
    }
    return ans;
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
         cin>>n>>m;
         cout<<(long long)((quickmod(m,n+1)-1)%mod*(quickmod(m-1,mod-2)%mod))%mod<<endl;
    }
    return 0;
}