HDU 4228 Flooring Tiles 反素数的应用

给你一个数N，找出一个最小的可以拆分成N种乘积表达形式的数x

比如N=2，6可以拆成2x3或者1x6两种，但不是最小的，最小的是4可以拆成1x4,2x2两种

首先可以肯定的是x必然有N*2或者是N*2-1（完全平方的情况）个约数

利用求反素数的过程求出约数为N*2和N*2-1个的最小的数

#include <cstdio>
#include <sstream>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <cctype>
#include <ctime>
#include <set>
#include <climits>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cmath>
#include <string>
#include <list>

#define INPUT_FILE "in.txt"
#define OUTPUT_FILE "out.txt"

using namespace std;

typedef unsigned long long LL;
const int INF = INT_MAX / 2;
const LL maxval = 1e18 + 1;
const int maxn = 75 + 5;
const int maxcnt = 160;
LL cnt[maxn * 2];
int prime[20] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71};
int times[20];

void dfs(LL curnum,LL curcnt,int nowt) {
    if(curnum > maxval) return;
    if(curcnt <= maxcnt) {
        cnt[curcnt] = min(cnt[curcnt],curnum);
    }
    for(int i = 1;i <= maxcnt;i++) {
        if(nowt == 0 || i <= times[nowt - 1]) {
            curnum *= prime[nowt]; 
            if(curnum > maxval) break;
            curcnt = curcnt / i * (i + 1);
            times[nowt] = i;
            dfs(curnum,curcnt,nowt + 1);
        }
        else break;
    }
}

int main() {
    for(int i = 1;i <= maxcnt;i++) cnt[i] = maxval;
    dfs(1,1,0);
    int N;
    while(cin >> N,N) {
        LL a1 = cnt[N * 2 - 1],a2 = cnt[N * 2],sa = sqrt(a1);
        if(sa * sa == a1 && a1 < a2) cout << a1 << endl;
        else cout << a2 << endl;
    }
    return 0;
}