Educational Codeforces Round 80 C. Two Arrays（组合数快速取模）

You are given two integers nn and mm . Calculate the number of pairs of arrays (a,b)(a,b) such that:

• the length of both arrays is equal to mm ;
• each element of each array is an integer between 11 and nn (inclusive);
• ai≤biai≤bi for any index ii from 11 to mm ;
• array aa is sorted in non-descending order;
• array bb is sorted in non-ascending order.

As the result can be very large, you should print it modulo 109+7109+7 .

Input

The only line contains two integers nn and mm (1≤n≤10001≤n≤1000 , 1≤m≤101≤m≤10 ).

Output

Print one integer – the number of arrays aa and bb satisfying the conditions described above modulo 109+7109+7 .

Examples

Input

2 2

Output

5

Input

10 1

Output

55

Input

723 9

Output

157557417

由题意可知，这个题所求的两个序列实际上可以合并成一个，将递增序列的头部接到递减序列的尾部，所得一个长为2m的序列。原问题可以转化为求范围1到n的2m个数组成的递增序列的个数（一个序列里可以有两个重复的数）。官方题解直接用Python+组合数公式

from math import factorial as fact
mod = 10**9 + 7

def C(n, k):
    return fact(n) // (fact(k) * fact(n - k))

n, m = map(int, input().split())
print(C(n + 2*m - 1, 2*m) % mod)

从原理入手，当这2m个数里有i个数不相同时，先从n个位置里用C(n,i)算出当前有几种选法，接下来有2m-i个数是和刚刚选出来的序列的部分数相同，可以看作同球入不同盒问题直接应用公式计算即可，每次循环更新ans。题目要求取模，这里借鉴了https://www.csdn.net/gather_2b/NtzaIgxsNzQtYmxvZwO0O0OO0O0O.html里组合数快速取模的知识。

#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
long long n,m;
int inv(int a)
{
    return a==1?1:(long long)(MOD-MOD/a)*inv(MOD%a)%MOD;
}
long long C(long long n,long long m)//组合数快速取模
{
    if(m<0)return 0;
    if(n<m)return 0;
    if(m>n-m)m=n-m;
    long long up=1,down=1;
    long long i;
    for(i=0;i<m;i++)
    {
        up=up*(n-i)%MOD;
        down=down*(i+1)%MOD;
    }
    return up*inv(down)%MOD;
}
int main()
{
    cin>>n>>m;
    int i;
    long long ans=0;
    for(i=2*m;i>=1;i--)//有i个不同的数
    {
        if(i>n)continue;
        ans=(ans+C(n,i)*/*同球入不同盒 2*m-i入i */ C(2*m-i+i-1,i-1))%MOD;
    }
    printf("%lld",ans);
    return 0;
}

原文地址：https://www.cnblogs.com/lipoicyclic/p/12246958.html