hdu3189-Just Do It-(埃氏筛+唯一分解定理)

Just Do It

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1552    Accepted Submission(s): 1036

Problem Description

Now we define a function F(x), which means the factors of x. In particular, F(1) = 1,since 1 only has 1 factor 1, F(9) = 3, since 9 has 3 factors 1, 3, 9. Now give you an integer k, please find out the minimum number x that makes F(x) = k.

 

Input

The first line contains an integer t means the number of test cases.
The follows t lines, each line contains an integer k. (0 < k <= 100).

 

Output

For each case, output the minimum number x in one line. If x is larger than 1000, just output -1.

 

Sample Input

4 4 2 5 92

 

Sample Output

6 2 16 -1

翻译：输入k，求因子数为k的数最小是哪个。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<vector>
#include<iostream>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
#define ll long long
const int maxx=1e6+5;

int cnt[maxx];///下标是自然数，内容是这个自然数有多少个因子
int ans[maxx];///下标是因子数，内容是含有这么多因子数最小是哪个数

void init()//埃氏筛打表
{
    memset(ans,-1,sizeof(ans));
    memset(cnt,0,sizeof(cnt));
    for(int i=1;i<maxx;i++)
    {
        for(int j=i;j<maxx;j=j+i)
        {
            cnt[j]++;
        }
        if( ans[ cnt[i] ]==-1 )///i从1开始，越变越大，因子数也会越来越多
            ans[ cnt[i] ]=i;///ans的下标 第一次遇见 没有遇见过的因子数目，就把i传进去，此时的i最小
    }
}

int main()///hdu3189
{
    init();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        if(ans[n]>1000)
            printf("-1
");
        else
            printf("%d
",ans[n]);
    }
    return 0;
}