首页 > 代码库 > hdu5108Alexandra and Prime Numbers（素数的性质）

hdu5108Alexandra and Prime Numbers（素数的性质）

2024-08-03 20:22:44   218人阅读

题目链接：

huangjing

思路：每一个数都可以表示成若干个素数的乘积，那么可以对N从2一直枚举到sqrt（N），然后对每个数都能除到不能取余为止，那么后面的合数就不会除了，所以最后得到的数就是最大的质因子，然后直接N/最大的质因子，还有就是N=1的时候没有存在的数  。

题目：

Alexandra and Prime Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 259    Accepted Submission(s): 94


Problem Description
Alexandra has a little brother. He is new to programming. One day he is solving the following problem: Given an positive integer N, judge whether N is prime.
The problem above is quite easy, so Alexandra gave him a new task: Given a positive integer N, find the minimal positive integer M, such that N/M is prime. If such M doesn‘t exist, output 0.
Help him!
 

Input
There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.
N≤1,000,000,000.
Number of cases with N>1,000,000 is no more than 100.
 

Output
For each case, output the requested M, or output 0 if no solution exists.
 

Sample Input
3
4
5
6
 

Sample Output
1
2
1
2
 

Source
BestCoder Round #19
 

Recommend
#include<cstdio>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;


vector<int>ans;
int n;

int main()
{
    while(~scanf("%d",&n))
    {
        int k=n;
        ans.clear();
        if(n==1) printf("0\n");
        else
        {
            for(int i=2;i*i<=n;i++)
            {
                while(n%i==0)
                {
                    ans.push_back(i);
                    n=n/i;
                }
            }
            if(n!=1)  ans.push_back(n);
            sort(ans.begin(),ans.end());
            printf("%d\n",k/ans.back());
        }
    }
    return 0;
}


代码：


hdu5108Alexandra and Prime Numbers（素数的性质）


联系
我们