AtCoder1974 いろはちゃんとマス目 / Iroha and a Grid （组合数学+阶乘逆元）

D - いろはちゃんとマス目 / Iroha and a Grid

Time Limit: 2 sec / Memory Limit: 256 MB

Score : 400400 points

Problem Statement

We have a large square grid with HH rows and WW 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 AA rows and the leftmost BB columns. (That is, there are A×BA×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+7109+7.

Constraints

• 1≦H,W≦100,0001≦H,W≦100,000
• 1≦A<H1≦A<H
• 1≦B<W1≦B<W

Input

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

HH WW AA BB

Output

Print the number of ways she can travel to the bottom-right cell, modulo 109+7109+7.

Sample Input 1 Copy

Copy

2 3 1 1

Sample Output 1 Copy

Copy

2

We have a 2×32×3 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

Copy

10 7 3 4

Sample Output 2 Copy

Copy

3570

There are 1212 forbidden cells.

Sample Input 3 Copy

Copy

100000 100000 99999 99999

Sample Output 3 Copy

Copy

1

Sample Input 4 Copy

Copy

100000 100000 44444 55555

Sample Output 4 Copy

Copy

738162020

题意：

给定H,W,A,B，其中H，W分别表示矩形的长宽，A,B表示左下角矩形不能通过，求从(1,1)到(H,W)的方案数，答案对1e9+7取模。

分析：

我们知道不带不可以走的左下角矩形的方案总数：C(H+W-2,H-1)。

错误的情况：（1，1）->I：C(H-A+i-2,H-A-1)

                          I->(H,W):   C(A-1+W-i,A-1)

 乘法原理：C(H-A+i-2,H-A-1)* C(A-1+W-i,A-1)

注意这里需要逆元：

C（n，m）=n！/m！*（n-m）！=n！*inv（m）*inv（n-m）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;     //  必须为质数才管用
const ll MAXN = 2e5 + 3;

ll fac[MAXN];       //  阶乘
ll inv[MAXN];       //  阶乘的逆元

ll QPow(ll x, ll n)
{
    ll ret = 1;
    ll tmp = x % MOD;
    while (n)
    {
        if (n & 1)
        {
            ret = (ret * tmp) % MOD;
        }
        tmp = tmp * tmp % MOD;
        n >>= 1;
    }

    return ret;
}

void Init()
{
    fac[0] = 1;
    for (int i = 1; i < MAXN; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
    }
    inv[MAXN - 1] = QPow(fac[MAXN - 1], MOD - 2);
    for (int i = MAXN - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % MOD;
    }
}
ll C(int a,int b) {
    if(a<0||b<0)return 1;
    return fac[a]*inv[b]%MOD*inv[a-b]%MOD;
}
int main() {
    int H,W,A,B;
    int N;
    scanf("%d%d%d%d",&H,&W,&A,&B);
    N=H+W-2;
    Init();
    ll ans=C(H+W-2,H-1);
    for(ll i=1;i<=B;i++) {
        ans-=C(H-A+i-2,H-A-1)*C(A-1+W-i,A-1)%MOD;
        ans=(ans%MOD+MOD)%MOD;
    }
    printf("%lld",ans);
    return 0;
}