HDU 2710 Max Factor 数论 (水

最新推荐文章于 2022-03-03 08:27:19 发布

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最新推荐文章于 2022-03-03 08:27:19 发布

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原文链接：http://www.cnblogs.com/Hyouka/p/5727599.html

本文介绍了一个编程问题：找出N个整数中拥有最大质因数的那个整数，并提供了完整的解决方案及代码实现。

Max Factor

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Description

To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the range 1..20,000. Unfortunately, he is unaware that the cows interpret some serial numbers as better than others. In particular, a cow whose serial number has the highest prime factor enjoys the highest social standing among all the other cows.

(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).

Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.

Input

* Line 1: A single integer, N

* Lines 2..N+1: The serial numbers to be tested, one per line

Output

* Line 1: The integer with the largest prime factor. If there are more than one, output the one that appears earliest in the input file.

Sample Input

Sample Output

38


比较水xjb做
新风格不错 以后就这样写了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define INF     0x3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;

const int MAXN=20005;

int p[MAXN];

int main()
{
    //FIN
    p[1]=1;
    int cou=0;
    for(int i=2; i<=20005; i++)
    {
        if(p[i]==0)
        {
            cou++;
            for(int j=i; j<=20005; j=j+i)
            {
                p[j]=cou;
            }
        }
    }
    int n;
    while(~scanf("%d",&n))
    {
        int ans=0,pos;
        int x;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            if(p[x]>ans)  {ans=p[x];pos=x;}
        }
        printf("%d\n",pos);

    }

}

转载于:https://www.cnblogs.com/Hyouka/p/5727599.html