[数论][组合数学]Iroha and a Grid

数论][组合数学]Iroha and a Grid"/>

[数论][组合数学]Iroha and a Grid

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 10 ⁹+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".

思路：总的走法减去错误走法；总的走法数为f((1,1)—>(h,w))（记f((a,b)—>(c,d))为从(a,b)走到(c,d)的走法数），错误的走法数为（公式含义易看出）

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
const long long mod=1e9+7;
typedef long long ll;
using namespace std;ll f[1000010],revf[1000010];//数组大小至少要为1e5*2ll 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;//防止出现负数tot=(tot-tmp)%mod;}cout<<tot<<endl;return 0;
}

 

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