Luogu5505 JSOI2011 points special product

Description

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There are \ (n \) classmates and \ (m \) specialty products.

Find the number of solutions

\(n,m\le10^3\)

Solution

Let's get started

Suppose we don't need to consider the situation without special products, directly upload the statistical method of the board method

\[f(i)=\prod^{m} _ {i=1}\binom{a_i+n-1}{n-1} \]

Then we consider this to subtract the situation where someone has no specialty

First, there is a man without deleting specialty situations, is given to \ (n-1 \) personal chant

Here we found that if directly \ (f_n-f_ {n-1} \) is obviously a fake approach

Because by definition, this \ (f_ {n-1} \) is likely to be assigned by \ (n-2 \) individuals, but not by \ (1 \) individuals

So we have to continue to rebuke

Another: Since we do n’t know which \ (i \) individual was not assigned, we still have to multiply it by one \ (\ binom {n} {ni} \)

Code

#include<bits/stdc++.h>
using namespace std;
#define int long long
namespace yspm{
	inline int read()
	{
		int res=0,f=1; char k;
		while(!isdigit(k=getchar())) if(k=='-') f=-1;
		while(isdigit(k)) res=res*10+k-'0',k=getchar();
		return res*f;
	}
	const int N=3e3+10,mod=1e9+7;
	int inv[N],fac[N];
	inline void prework()
	{
		fac[0]=1; inv[0]=1; inv[1]=1;
		for(int i=1;i<N;++i) fac[i]=fac[i-1]*i%mod;
		for(int i=2;i<N;++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		for(int i=1;i<N;++i) inv[i]*=inv[i-1],inv[i]%=mod;
		return ; 
	}
	inline int C(int n,int m)
	{
		if(n<0||m<0||n<m) return 0; 
		return inv[n-m]*inv[m]%mod*fac[n]%mod;
	}
	int a[N],n,m,ans,f[N];
	signed main()
	{ 
		prework();
		n=read(); m=read(); 
		for(int i=1;i<=m;++i) a[i]=read();
		for(int i=1;i<=n;++i)
		{
			f[i]=1;
			for(int j=1;j<=m;++j) f[i]*=C(a[j]+i-1,i-1),f[i]%=mod;
		}
		for(int i=0;i<=n;++i)
		{
			if(i&1)
			{
				ans-=C(n,i)*f[n-i];
				ans+=mod; ans%=mod;
			}
			else 
			{
				ans+=C(n,i)*f[n-i]%mod;
				ans%=mod;
			}
		}cout<<ans<<endl;
		return 0;
	}
}
signed main(){return yspm::main();}

Review

When we solve the problem of tolerance, we need to use a variety of combinations

At the same time, accurately find duplicate information and then enumerate and repel