题意:
有一个n * m的矩形,一开始从(1, 1)开始,每次能够到达右下方格子的一个格子,求到达(n, m)的方案数。
题解:
很显然最多走min (n - 2, m - 2) + 1步,枚举步数K,答案为 求和 C(n - 2, K - 1) * C(m - 2, K - 1)
代码:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e5 + 7;
#define LL long long
int n, m;
LL F[N], ans;
map <int, LL> inv;
int Pow (int x, int cnt) {
int ret = 1;
while (cnt) {
if (cnt & 1) ret = (LL) ret * x % mod;
x = (LL) x * x % mod;
cnt >>= 1;
}
return ret;
}
int main () {
scanf ("%d%d", &n, &m);
F[0] = 1, inv[0] = 1;
for (int i = 1; i < N; ++i) {
F[i] = F[i-1] * i % mod;
inv[i] = Pow (F[i], mod - 2);
}
int lim = min (n - 2, m - 2);
for (int i = 1; i <= lim + 1; ++i) {
LL ans1 = F[n-2] * inv[i-1] % mod * inv[n-i-1] % mod;
LL ans2 = F[m-2] * inv[i-1] % mod * inv[m-i-1] % mod;
ans = (ans + ans1 * ans2 % mod) % mod;
}
cout << ans << endl;
return 0;
}
原文:http://www.cnblogs.com/xgtao/p/6024549.html