51Nod－1119－机器人走方格 V2

描述

题解

C(n - 1 + m - 1, n - 1)

代码

#include<iostream>
#include<algorithm>
#include<cmath>

using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;

//  返回d = gcd(a,b); 和对应于等式ax + by = d中的x,y
ll extend_gcd(ll a, ll b, ll &x, ll &y)
{
if (a == 0 && b == 0)
{
return -1;  //  无最大公约数
}
if (b == 0)
{
x = 1;
y = 0;
return a;
}
ll d = extend_gcd(b, a % b, y, x);
y -= a / b * x;
return d;
}

//  求逆元素
//  ax = 1(mod n)
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 m, ll n)
{
ll i, t_1, t_2;
t_1 = t_2 = 1;
for (i = n; i >= n - m + 1; i--)
{
t_1 = t_1 * i % mod;
}
for (i = 1; i <= m; i++)
{
t_2 = t_2 * i % mod;
}
return  t_1 * mod_reverse(t_2, mod) % mod;  //  转换为逆元
}

int main()
{
ll n, m, ans;
cin >> m >> n;
ans = c(min(m - 1, n - 1), m + n - 2);
cout << ans << endl;
return 0;
}

参考