bzoj 1025 [SCOI2009]游戏 dp

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;
}