bzoj 4714: 旋转排列

最新推荐文章于 2023-06-19 14:36:47 发布

ypxrain

最新推荐文章于 2023-06-19 14:36:47 发布

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分类专栏：组合数学

本文链接：https://blog.youkuaiyun.com/ypxrain/article/details/80099613

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这里写图片描述
1≤n≤500000

Solution

我们设 $f_i$ 表示长度为i的错排，容斥后易得答案

代码

#include <bits/stdc++.h>

typedef long long LL;

const int N = 500005;
const int MOD = 1000000007;

int n,jc[N],ny[N],f[N];

int C(int n,int m)
{
    return (LL)jc[n] * ny[m] % MOD * ny[n - m] % MOD;
}

int main()
{
    scanf("%d",&n);
    jc[0] = jc[1] = ny[0] = ny[1] = 1;
    for (int i = 2; i <= n; i++) 
        jc[i] = (LL)jc[i - 1] * i % MOD, ny[i] = (LL)(MOD - MOD / i) * ny[MOD % i] % MOD;
    for (int i = 2; i <= n; i++) 
        ny[i] = (LL)ny[i - 1] * ny[i] % MOD;
    f[0] = f[2] = 1;
    for (int i = 3; i <= n; i++) 
        f[i] = (LL)(i - 1) * (f[i - 1] + f[i - 2]) % MOD;
    int ans = 0;
    for (int i = 2; i <= n; i++)
    {
        int w = 1;
        for (int j = 1; j * i <= n; j++)
        {
            w = (LL)w * C(n - i * (j - 1), i) % MOD * jc[i - 1] % MOD;
            if (j & 1) 
                (ans += (LL)w * f[n - i * j] % MOD * ny[j] % MOD) %= MOD;
            else (ans += MOD - (LL)w * f[n - i * j] % MOD * ny[j] % MOD) %= MOD;
        }
    }
    printf("%d",ans);
}