[USACO 09FEB]Bullcow

Description

題庫連接

有 \(n\) 頭牛，每頭牛能夠爲 \(\text{A}\) 牛也能夠爲 \(\text{B}\) 牛。如今給這些牛排隊，要求相鄰兩頭 \(\text{A}\) 牛之間至少間隔 \(k\) 頭 \(\text{B}\) 牛，求方案數，對大質數取模。

\(0\leq k<n\leq 100000\)

Solution

考慮枚舉有幾頭 \(\text{A}\) 牛，設爲 \(i\)。

\(\text{B}\) 牛數爲 \(n-i\) 。由墊球法以及隔板法，可知當前狀況下方案爲

\[{n-i-(i-1)\times k+i}\choose i\]

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;   
}