題目描述

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

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

輸入

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‘.

輸出

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

樣例輸入

2
16
19
0

樣例輸出

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：當前枚舉到的數；
    //cnt：該數的因數個數；
    //limit：因數個數的上限；2^t1*3^t2*5^t3……t1>=t2>=t3……
    //第k大的素數
    if(cur>(1LL<<60) || cnt>150) return ;
    if(p[cnt]!=0 && p[cnt]>cur)//當前的因數個數已經記錄過且當時記錄的數比當前枚舉到的數要大，則替換此因數個數下的枚舉到的數
        p[cnt]=cur;
    if(p[cnt]==0)//此因數個數的數還沒有出現過，則記錄
        p[cnt]=cur;
    LL temp=cur;
    for(int i=1; i<=limit; i++) //枚舉數
    {
        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\n", p[n]);
    }
}

HDUSTOJ-1558 Flooring Tiles（反素數）