4710: [Jsoi2011]分特产

4710: [Jsoi2011]分特产

链接

分析：

　　容斥原理+隔板法。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 2005, mod = 1e9 + 7;
int C[N][N], a[N], n, m;

int Calc(int x) {
    int ans = 1;
    for (int i = 1; i <= m; ++i) 
        ans = 1ll * ans * C[a[i] + x - 1][x - 1] % mod;
    ans = 1ll * ans * C[n][x] % mod;
    return ans;
}

int main() {
    n = read(), m = read();
    for (int i = 1; i <= m; ++i) a[i] = read();
    C[0][0] = 1;
    for (int i = 1; i <= 2000; ++i) { 
        C[i][0] = 1;
        for (int j = 1; j <= i; ++j) 
            C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;    
    }
    int ans = Calc(n);
    for (int i = 1; i <= n; ++i) {
        if (i & 1) ans = (ans - Calc(n - i) + mod) % mod;
        else ans = (ans + Calc(n - i)) % mod;        
    }
    cout << (ans + mod) % mod;
    return 0;
}