[USACO 09FEB]Bullcow
Description
有 $n$ 头牛,每头牛可以为 $\text{A}$ 牛也可以为 $\text{B}$ 牛。现在给这些牛排队,要求相邻两头 $\text{A}$ 牛之间至少间隔 $k$ 头 $\text{B}$ 牛,求方案数,对大质数取模。
$0\leq k<n\leq 100000$
Solution
考虑枚举有几头 $\text{A}$ 牛,设为 $i$。
$\text{B}$ 牛数为 $n-i$ 。由垫球法以及隔板法,可知当前情况下方案为
$${n-i-(i-1)\times k+i}\choose i$$
Code
#include <bits/stdc++.h>
using namespace std;
const int N = 200000+5, yzh = 5000011;
int fac[N], ifac[N], n, k, ans = 1;
int C(int n, int m) {return 1ll*fac[n]*ifac[m]%yzh*ifac[n-m]%yzh; }
int main() {
scanf("%d%d", &n, &k);
fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
for (int i = 2; i <= (n<<1); i++)
fac[i] = 1ll*i*fac[i-1]%yzh;
for (int i = 2; i <= (n<<1); i++)
ifac[i] = -1ll*yzh/i*ifac[yzh%i]%yzh;
for (int i = 2; i <= (n<<1); i++)
ifac[i] = 1ll*ifac[i]*ifac[i-1]%yzh;
for (int i = 1; i <= n && n-i-(i-1)*k >= 0; i++)
(ans += C(n-i-(i-1)*k+i, i)) %= yzh;
printf("%d\n", (ans+yzh)%yzh);
return 0;
}
博主蒟蒻,随意转载。但必须附上原文链接:http://www.cnblogs.com/NaVi-Awson/,否则你会终生找不到妹子!!!