Iroha and a Grid(组合数学，逆元)

Iroha and a Grid

时间限制: 1 Sec   内存限制: 128 MB

题目描述

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

输入

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

H W A B

输出

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

样例输入

2 3 1 1

样例输出

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".

来个超时的代码，实际上应该打一个N！的表

#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
typedef long long ll;

ll x,y;

ll extgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0){
        x = 1;
        y = 0;
        return a;
    }
    ll gcd = extgcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a/b) * y;
    return gcd;
}

ll J(ll n,ll x)
{
    ll ans = 1;
    for(ll i=x;i<=n;i++)
        ans = (ans*i)%mod;
    return ans;
}

ll C(ll m,ll n)
{
    extgcd(J(m,2),mod,x,y);
    return J(n,n-m+1)*(x+mod)%mod;
}

int main()
{
    ll h,w,A,B;
    cin>>h>>w>>A>>B;

    ll ans=0;

    for(ll i=B+1;i<=w;i++)
    {
        ll tmp=(C(h-A-1,i-1+h-A-1)*C(w-i,w-i+A-1))%mod;
        ans=(ans+tmp)%mod;
    }
    cout<<ans<<endl;
    return 0;
}

打表后AC

#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
typedef long long ll;

ll JE[200010];

ll x,y;

ll extgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0){
        x = 1;
        y = 0;
        return a;
    }
    ll gcd = extgcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a/b) * y;
    return gcd;
}

ll C(ll m,ll n)
{
    extgcd(JE[m]*JE[n-m]%mod,mod,x,y);
    return JE[n]*(x+mod)%mod;
}

int main()
{
    ll h,w,A,B;

    JE[0]=1;

    for(ll i=1;i<200005;i++)
        JE[i]=(JE[i-1]*i)%mod;

    cin>>h>>w>>A>>B;

    ll ans=0;

    for(ll i=B+1;i<=w;i++)
    {
        ll tmp=(C(h-A-1,i-1+h-A-1)*C(w-i,w-i+A-1))%mod;
        ans=(ans+tmp)%mod;
    }
    cout<<ans<<endl;
    return 0;
}