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解法

顯然，可以回到初始狀態就意味著一定由若幹個環組成
假設環的長度為\(l_i\)
那麽，我們可以得到\(\sum l_i=n\)
不考慮自環的情況，那麽\(\sum l_i≤n\)
將\(n\)以內的質因數全部篩出，強制每一次只取某一個質數的次冪，那麽就可以解決重復計數的問題
時間復雜度：\(O(n^2)\)

代碼

#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;
}
 

bzoj 1025 [SCOI2009]遊戲 dp