【组合数】51Nod 1627 瞬间移动

题面在这里

考虑枚举移动的步数$i$

那么答案就是

$$\sum_{i=1}^{i<min(n,m)}C_{n-2}^{i-1}\cdot C_{m-2}^{i-1}$$

示例程序：

#include<cstdio>
typedef long long ll;

const int maxn=100005,tt=1e9+7;
int n,m;
ll fac[maxn],inv[maxn],ans;
ll power(ll a,int b){
    ll w=a,res=1;
    while (b){
        if (b&1) res=res*w%tt;
        w=w*w%tt;
        b>>=1;
    }
    return res;
}
void make(){
    fac[0]=1;inv[0]=1;
    for (int i=1;i<=100000;i++) fac[i]=fac[i-1]*i%tt,inv[i]=power(fac[i],tt-2);
}
inline ll C(int y,int x){
    return fac[x]*inv[y]%tt*inv[x-y]%tt;
}
int main(){
    scanf("%d%d",&n,&m);
    if (n==2||m==2) return printf("1"),0;
    make();ans=0;
    for (int i=1;i<n&&i<m;i++)
     (ans+=C(i-1,n-2)*C(i-1,m-2)%tt)%=tt;
    printf("%lld",ans);
    return 0;
}