bzoj4710 分特产

嘟嘟嘟


因为刚刚做过一道类似的，所以感觉容斥可做。


但自己还是没有搞出来，容斥这方面没做多少题果然是硬伤。
对于每一种特产都用插板法算出分配方案，然后乘起来就是有人可能啥都没得到的方案数，即\(\prod C_{n + a_i + 1} ^ {n - 1}\)。
这时候用容斥，减去至少有一个人什么都没有，加上至少有两个人什么都没有，减去三个人……然后就没了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e3 + 5;
const ll mod = 1e9 + 7;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
#endif
}

int n, m, a[maxn];

ll fac[maxn], inv[maxn];
In ll inc(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}
In ll C(int n, int m) {return fac[n] * inv[m] % mod * inv[n - m] % mod;}
In ll quickpow(ll a, ll b)
{
  ll ret = 1;
  for(; b; b >>= 1, a = a * a % mod)
    if(b & 1) ret = ret * a % mod;
  return ret;
}

In void init()
{
  fac[0] = inv[0] = 1;
  for(int i = 1; i < maxn; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[maxn - 1] = quickpow(fac[maxn -  1], mod - 2);
  for(int i = maxn - 2; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
}

int main()
{
  //MYFILE();
  n = read(), m = read();
  for(int i = 1; i <= m; ++i) a[i] = read();
  init();
  ll ans = 0;
  for(int i = n; i >= 0; --i)
    {
      ll tp = C(n, n - i);
      for(int j = 1; j <= m; ++j) tp = tp * C(i + a[j] - 1, i - 1) % mod;
      ans = inc(ans, ((n - i) & 1) ? mod - tp : tp);
    }
  write(ans), enter;
  return 0;
}