题面

题目传送门

解法

显然，可以回到初始状态就意味着一定由若干个环组成

假设环的长度为\(l_i\)

那么，我们可以得到\(\sum l_i=n\)

不考虑自环的情况，那么\(\sum l_i≤n\)

将\(n\)以内的质因数全部筛出，强制每一次只取某一个质数的次幂，那么就可以解决重复计数的问题

时间复杂度：\(O(n^2)\)

代码

#include <bits/stdc++.h>
#define int long long
#define N 1010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == ‘-‘) f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - ‘0‘, c = getchar(); x *= f;
}
int len, a[N], f[N][N];
bool prime(int x) {
    for (int i = 2; i * i <= x; i++)
        if (x % i == 0) return false;
    return true;
}
void sieve(int n) {
    len = 0;
    for (int i = 2; i <= n; i++)
        if (prime(i)) a[++len] = i;
}
main() {
    int n; read(n);
    sieve(n); f[0][0] = 1;
    for (int i = 1; i <= len; i++)
        for (int j = 0; j <= n; j++) {
            f[i][j] += f[i - 1][j];
            for (int k = a[i]; j + k <= n; k = k * a[i])
                f[i][j + k] += f[i - 1][j];
        }
    int ans = 0;
    for (int i = 0; i <= n; i++) ans += f[len][i];
    cout << ans << "\n";
    return 0;
}

原文地址：https://www.cnblogs.com/copperoxide/p/9472318.html