hdu5793 A Boring Question（推公式or迷之找规律）-优快云博客

本文探讨了一个复杂的组合数学问题，通过巧妙的数学推导找到了一个简洁的公式来解决该问题。该公式涉及到组合数的计算，并最终简化为一个关于m和n的幂运算表达式。

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思路：比赛时候是迷之手玩了很多个样例然后大胆猜测了一发公式...

正解：

A Boring Question

$\sum_{0\leq k_{1},k_{2},\cdots k_{m}\leq n}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}}$ $=\sum_{0\leq k_{1}\leq k_{2}\leq\cdots \leq k_{m}\leq n}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}}$ $=\sum_{k_{m}=0}^{n}\sum_{k_{m-1}=0}^{k_{m}}\cdots \sum_{k_{1}=0}^{k_{2}}\prod_{1\leq j< m}\binom{k_{j+1}}{k_{j}}$ $=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{1}=0}^{k_{2}}\binom{k_{2}}{k_{1}} \right } \right }$ $=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{1}=0}^{k_{2}}\binom{k_{2}}{k_{1}} \right } \right }$ $=\sum_{k_{m}=0}^{n}\left { \binom{k_{m}}{k_{m-1}} \sum_{k_{m-1}=0}^{k_{m}} \left { \binom{k_{m-1}}{k_{m-2}} \cdots \sum_{k_{2}=0}^{k_{3}}\binom{k_{3}}{k_{2}}2^{k_{2}} \right } \right }$ $=\sum_{k_{m}=0}^{n}m^{k_{m}}$ $=\frac{m^{n+1} - 1}{m - 1}$

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
#define LL long long

template<class T1,class T2>
T1 quick_mod(T1 t,T2 n)
{
	T1 ans=1;
	while(n)
	{
		if(n&1) ans=ans*t%mod;
		t=t*t%mod;
		n>>=1;
	}
	return ans;
}
LL Inv(LL x)
{
	return quick_mod(x, mod-2);
}
int main()
{
    int T;
	scanf("%d",&T);
	while(T--)
	{
		LL n,m;
		scanf("%lld%lld",&n,&m);
		if(n==0)
		{
			printf("1\n");
			continue;
		}
		LL inv = Inv(m-1);
		LL ans = ((quick_mod(m,n+1)-1)*inv)%mod;
		printf("%lld\n",ans);
	}
}

Problem Description

There are an equation.