ever nts constrain names 公式 bottom stand and 數論

題目描述

We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell.
However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells.
Find the number of ways she can travel to the bottom-right cell.
Since this number can be extremely large, print the number modulo 109+7.

Constraints
1≤H,W≤100,000
1≤A<H
1≤B<W

輸入

The input is given from Standard Input in the following format:

H W A B

輸出

Print the number of ways she can travel to the bottom-right cell, modulo 10⁹+7.

樣例輸入

2 3 1 1

樣例輸出

2

提示

We have a 2×3 grid, but entering the bottom-left cell is forbidden. The number of ways to travel is two: "Right, Right, Down" and "Right, Down, Right".

思路：總的走法減去錯誤走法；總的走法數為f((1,1)—>(h,w))（記f((a,b)—>(c,d))為從(a,b)走到(c,d)的走法數），錯誤的走法數為（公式含義易看出）

AC代碼：

#include <iostream>
#include<cstdio>
#include<algorithm>
const long long mod=1e9+7;
typedef long long ll;
using namespace std;

ll f[1000010],revf[1000010];//數組大小至少要為1e5*2

ll qpow(ll a,ll b){
  ll ret=1;
  while(b){
    if(b&1) ret=(ret*a)%mod;
    a=(a*a)%mod;
    b>>=1;
  }
  return ret;
}

void init(){
  f[0]=1; revf[0]=qpow(f[0],mod-2);
  for(ll i=1;i<1000010;i++){
    f[i]=i*f[i-1]%mod;
    revf[i]=qpow(f[i],mod-2);
  }
}

ll C(ll n,ll m){
  return (f[n]*revf[m])%mod*revf[n-m]%mod;
}

ll count_ways(ll a,ll b,ll c,ll d){
  ll tot=(c-a)+(d-b);
  ll down=(c-a);
  ll ret=C(tot,down);
  return ret;
}

int main()
{
    init();
    ll h,w,a,b;
    cin>>h>>w>>a>>b;
    ll tot=count_ways(1,1,h,w);
    for(ll i=1;i<=b;i++){
        ll tmp=count_ways(1,1,h-a,i)*count_ways(h-a+1,i,h,w)%mod;
        while(tot<tmp) tot+=mod;//防止出現負數
        tot=(tot-tmp)%mod;
    }
    cout<<tot<<endl;
    return 0;
}

[數論][組合數學]Iroha and a Grid