CF768F Barrels and boxes

Toot toot


this question is not difficult.
Such questions do know a few of these routines: we enumerate several piles of wine, so you can calculate how much heap of food as well as their number of permutations, then the probability is that the number of legal program / total program.
Provided wine \ (I \) stack, then there is \ (C_ {w - 1} ^ {i - 1} \) permutations method, the number corresponding to the heap of food may have \ (i - 1, i, i 1 + \) stack, and the food arrangement method of calculating the same separator method, i.e. \ (C_ {f - 1} ^ {i - 2}, C_ {f - 1} ^ {i - 1}, C_ {f - I. 1} ^ {} \) . Take these two numbers together is the wine for the heap when \ (i \) of the total program.
As for the number of legal program, is forced to let a bunch of wine put on each \ (h \) months, then go \ (i \) heap put on wine \ (w - h * i \ ) , namely \ (C_ W {- H * I - I. 1} ^ {-}. 1 \) .
Then pay attention to the border situation.

#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;
}