【bzoj 1025】游戏（DP）

传送门biu~
题目可以转化为每一个对应关系a-b为从a->b的一条边，求存在的所有环的大小的最小公倍数有几种情况。考虑dp，只考虑质数的情况可以分解质因数，f[i][j]代表选前i个质数的和不超过j的情况数。

#include<bits/stdc++.h>using namespace std;int n;long long ans,f[1005][1005];int prime[]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};int main(){    scanf("%d",&n);    f[0][0]=1;    for(int i=1;i<=168;++i){        for(int j=0;j<=n;++j)   f[i][j]=f[i-1][j];        for(int j=prime[i];j<=n;j*=prime[i]){            for(int k=0;k+j<=n;++k){                f[i][k+j]+=f[i-1][k];            }        }    }    for(int i=0;i<=n;++i)   ans+=f[168][i];    printf("%lld",ans);    return 0;}