いろはちゃんとマス目 / Iroha and a Grid（AtCoder-1974）

Problem Description

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.

Example

Sample Input 1

2 3 1 1

Sample Output 1

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

10 7 3 4

Sample Output 2

3570
There are 12 forbidden cells.

Sample Input 3

100000 100000 99999 99999

Sample Output 3

1

Sample Input 4

100000 100000 44444 55555

Sample Output 4

738162020

题意：有一个区域，大小是 h*w，现在有一个 (1,1) 到 (A,B) 的禁区不能走，现在要在左上角 (1,h) 走到右下角 (w,1)，每次只能向右或向下走，问有多少种走法，最后答案对 1E9+7 取模

思路：

首先考虑无禁区的情况：从 (1,h) 走到 (w,1) 总共走 h+w-2 步，而在走到最高一行前，每次都可以选择是否向下走，即可以选择 h-1 次，那么方案数即为 ，

再考虑有禁区的情况：

每个合法路径都会到达第 h-A 行的 B 之后的某一列上，则每个合法的路径都可以分为从 (1,1) 走到第 h-A 行 B 之后的第 i 列上，与第 h-A 行 B 之后的第 i 列走到 (h,w) 上，然后考虑第 h-A 行 B 之后的第 i 列不能向右移动，那么总的路径就变为 3 段，即：，其中，i 从 B+1 到 m

再考虑取模的，由于 1E9+7 是个素数，因此可以考虑使用逆元的方法来求，要注意的是，由于 h 与 w 最大为 1E5，那么预处理阶乘时最大要处理到 w+h，即 2E5。

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 200000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;
LL factor[N];
LL PowMod(LL a,LL b){
    LL res=1;
    while(b){
        if(b&1)
            res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
LL inv(LL x){
    return PowMod(x,MOD-2);
}
LL C(LL n,LL m){
    return factor[n]*inv(factor[m])%MOD*inv(factor[n-m])%MOD;
}
void init(){
    factor[0]=1;
    for(int i=1;i<=N;i++)
        factor[i]=(factor[i-1]*i)%MOD;
}
int main(){
    init();
    LL h,w,A,B;
    scanf("%lld%lld%lld%lld",&h,&w,&A,&B);
    LL res=0;
    for(LL i=B+1;i<=w;i++){
        LL temp=(C(i-1+h-A-1,h-A-1)*C(w-i+A-1,w-i))%MOD;
        res=(res+temp)%MOD;
    }
    printf("%lld\n",res);
    return 0;
}