ZOJ 2562 More Divisors 反素数 DFS

//素数表 prime{0,2,3,5,7,11,13,17,19};
//如果一个数num = prime ^ (2 * n - 1) (基数的情况) 则它的因子数为 2 * n ;
//如果一个数num = prime ^ (2 * n    ) (偶数的情况) 则它的因子数为 2 * n + 1; (prime 是一个质素);
//x = prime[1]^( k1 - 1 ) * prime[2] * ( k2 - 1 ) * prime[3] * ( k3 - 1 ) * ... * prime
* ( kn - 1 );
//因子数为k , k = k1 * k2 * k3 * ... * kn;
//找出比 N 小的 最小的 x 使得 k 最大;
//k1 >= k2 >= k3 >= ... >= kn;

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

long long prime[20] = {0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
long long n;
long long INF = (long long) 1 << 60;

long long numk;
long long totn ;

long long fast_pow(int x,int k) {
long long base = x;
long long ans = 1;
while(k) {
if(k & 1) ans *= base;
base *= base;
k >>= 1;
}
return ans;
}

void DFS(int depth,long long num,int lim ,long long tot) {
//depth 搜素深度
//num 因子数
//lim 是下个因子的最大数目
//tot 数的大小
//printf("depth : %d num : %lld tot : %lld\n",depth,num,tot);
//printf("totn  : %lld numk : %lld\n",totn,numk);
if(tot > n) return ;
if(tot <= 0) return ;
if(num == numk && tot < totn) {
totn = tot;
}
if(num > numk) {
totn = tot;
numk = num;
}
for(int i = 2 ; i <= lim; i++) {
//枚举每层k值得大小;
DFS(depth+1,num * i, i ,tot * fast_pow(prime[depth],i-1) );
}
}

void Deal_with() {
while(~scanf("%lld",&n)) {
numk = 0; totn = INF;
//printf("totn  : %lld numk : %lld\n",totn,numk);
DFS(1,1,50,1);
printf("%lld\n",totn);
}
}

int main(void) {
//freopen("a.in","r",stdin);
Deal_with();
return 0;
}