D - いろはちゃんとマス目 / Iroha and a Grid

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

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

Input

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

H W A B

Output

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

Sample Input 1

Copy

2 3 1 1

Sample Output 1

Copy

2

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".

Sample Input 2

Copy

10 7 3 4

Sample Output 2

Copy

3570

There are 12 forbidden cells.

Sample Input 3

Copy

100000 100000 99999 99999

Sample Output 3

Copy

1

Sample Input 4

Copy

100000 100000 44444 55555

Sample Output 4

Copy

738162020

解析：        看了這篇部落格我覺得已經不用再說什麼了。

程式碼：

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int mod=1e9+7;
int n,m,a,b,ans,mul[1000010];

inline void pre()
{
	mul[0]=1;
	for(int i=1;i<=1000000;i++) mul[i]=(mul[i-1]*i)%mod;
}

inline int ksm(int a,int b)
{
	int ans=1;
	a%=mod;
	while(b)
	{
	  if(b&1) ans=(ans*a)%mod;
	  b>>=1;
	  a=(a*a)%mod;
	}
	return ans;
}

inline int C(int n,int m)
{
	if(n>=0 && m>=0 && n>=m) return (mul[n]*ksm(mul[m],mod-2)%mod*ksm(mul[n-m],mod-2)%mod)%mod;
	return 0;
}

signed main()
{
	pre();
	scanf("%d%d%d%d",&n,&m,&a,&b);
	ans=C(n-1+m-1,n-1);
	for(int j=1;j<=b;j++)
	{
	  int i=n-a,sum=C(i-1+j-1,i-1)*C(n-i-1+m-j,n-i-1)%mod;
	  ans=((ans-sum)%mod+mod)%mod; 
	}
	cout<<ans<<"\n";
	return 0;
}