HDUSTOJ-1558 Flooring Tiles（反素数）

1558: Flooring Tiles

时间限制: 3 Sec  内存限制: 128 MB
提交: 59  解决: 36
[提交][状态][讨论版]

题目描述

You want to decorate your floor with square tiles. You like rectangles. With six square flooring tiles, you can form exactly two unique rectangles that use all of the tiles: 1 x 6, and 2 x 3 (6 x 1 is considered the same as 1 x 6. Likewise, 3 x 2 is the same as 2 x 3). You can also form exactly two unique rectangles with four square tiles, using all of the tiles: 1 x 4, and 2 x 2.

Given an integer N, what is the smallest number of square tiles needed to be able to make exactly N unique rectangles, and no more, using all of the tiles? If N = 2, the answer is 4.

输入

There will be several test cases in the input. Each test case will consist of a single line containing a single integer N,  which represents the number of desired rectangles. The input will end with a line with a single `0'.

输出

For each test case, output a single integer on its own line, representing the smallest number of square flooring tiles needed to be able to form exactly N rectangles, and no more, using all of the tiles. The answer is guaranteed to be at most 1018. Output no extra spaces, and do not separate answers with blank lines.

样例输入

2
16
19
0

样例输出

4
840
786432

#include<iostream>
#include<cstring>
#include<cstdio>
 
using namespace std;
typedef long long LL;
LL p[1010];
LL prime[30]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
void getartprime(LL cur,int cnt,int limit,int k)
{
    //cur：当前枚举到的数；
    //cnt：该数的因数个数；
    //limit：因数个数的上限；2^t1*3^t2*5^t3……t1>=t2>=t3……
    //第k大的素数
    if(cur>(1LL<<60) || cnt>150) return ;
    if(p[cnt]!=0 && p[cnt]>cur)//当前的因数个数已经记录过且当时记录的数比当前枚举到的数要大，则替换此因数个数下的枚举到的数
        p[cnt]=cur;
    if(p[cnt]==0)//此因数个数的数还没有出现过，则记录
        p[cnt]=cur;
    LL temp=cur;
    for(int i=1; i<=limit; i++) //枚举数
    {
        temp=temp*prime[k];
        if(temp>(1LL<<60)) return;
        getartprime(temp,cnt*(i+1),i,k+1);
    }
}
 
void Init(){
    getartprime(1, 1, 75, 0);
    for(int i = 1; i <= 75; i++){
        if(p[2*i] != 0 && p[2*i-1] != 0) p[i] = min(p[2*i], p[2*i-1]);
        else if(p[2*i] != 0) p[i] = p[2*i];
        else p[i] = p[2*i - 1];
    }
 
}
int main(){
    int n;
    Init();
    while(scanf("%d", &n) == 1 && n){
        printf("%lld
", p[n]);
    }
}