机器人每次只能向右走或者向下走。假设机器人在一个5*5的格子内部,机器人在格子内移动,每次只能走一格子,从左上走到右下,这样机器人会向右移动4次,向下移动4次,这就是4个向右移动和4个向下移动的排列组合,一共移动了8次,移动的方案数就是C(8,4)。由于求组合数的时候数字太大,而且也要取模,所以求组合数取模的时候要用乘法逆元
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
ll m,n;
const ll mod = 1e9+7;
ll extend_gcd(ll a, ll b, ll &x, ll &y)
{
if(a==0&&b==0)
return-1;
if(b==0)
{
x = 1;
y = 0;
return a;
}
ll d = extend_gcd(b,a%b,y,x);
y -= a/b*x;
return d;
}
ll mod_reverse(ll a, ll n)
{
ll x,y;
ll d = extend_gcd(a,n,x,y);
if(d == 1) return (x%n+n)%n;
else return -1;
}
ll c(ll n, ll m)
{
ll t1 = 1;
ll t2 = 1;
//分子
for(ll i = n; i >= m+1 ; --i)
t1 = t1*i%mod;
//分母
for(ll i = 1; i <= (n-m); ++i)
t2 = t2*i%mod;
return t1*mod_reverse(t2,mod)%mod;
}
int main()
{
cin >> m >> n;
cout << c(m+n-2,min(m-1,n-1))%mod << endl;
return 0;
}