CF768F Barrels and boxes

嘟嘟嘟


此题不难。
这种题做几道就知道些套路了：我们枚举酒有几堆，这样就能算出食物有多少堆以及他们的排列数，那么概率就是合法方案数 / 总方案数。
设酒有\(i\)堆，那么就有\(C_{w - 1} ^ {i - 1}\)种排列方法，对应的食物堆数就可能有\(i - 1, i, i + 1\)堆，然后同样用隔板法算出食物的排列方法，即\(C_{f - 1} ^ {i - 2}, C_{f - 1} ^ {i - 1}, C_{f - 1} ^ {i}\)。把这俩乘起来就是当酒堆数为\(i\)的总方案数。
至于合法方案数，就是我们先强制往每一堆酒上放\(h\)个，然后再往\(i\)堆酒上放\(w - h * i\)个，即\(C_{w - h * i - 1} ^ {i - 1}\)。
然后注意边界情况。

#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 = 1e5 + 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, h;

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)
{
  if(m < 0 || m > n) return 0;
  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();
  m = read(), n = read(), h = read();
  if(!m) {puts(n > h ? "1" : "0"); return 0;}
  if(!n) {puts("1"); return 0;}
  init();
  ll ans1 = 0, ans2 = 0;
  for(int i = 1; i <= n; ++i)
    {
      ll tp1 = C(n - 1, i - 1);
      ll tp2 = inc(C(m - 1, i - 2), inc(C(m - 1, i - 1) * 2, C(m - 1, i)));
      if(1LL * n - 1LL * h * i >= i) ans2 = inc(ans2, C(n - h * i - 1, i - 1) * tp2 % mod);
      ans1 = inc(ans1, tp1 * tp2 % mod);
    }
  write(ans2 * quickpow(ans1, mod - 2) % mod), enter;
  return 0;
}