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HDU 2710 Max Factor （筛选求素数）

2024-08-07 06:14:18   216人阅读

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3995    Accepted Submission(s): 1301


Problem 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
4
36
38
40
42
 

Sample Output
38

题目大意：给你N个数，求这N个数中哪个数的最大素因子最大，输出这个数，如果有多种结果，输出靠前边的那个数。

本例写的代码中prime具有二义性，但是这样能够减少空间利用

#include<iostream>
using namespace std;
int prime[20010];
int getprime()
{
    prime[1]=1;
    for(int i=2;i<20010;++i)
    {
        if(prime[i]==0)
        {
            for(int j=i;j<20010;j+=i)
            {
                prime[j]=i;
            }
        }
    }
}
int main(int argc, char *argv[])
{
    //freopen("2710.in", "r",stdin);
    int MAX=0;
    getprime();
    int n;
    int a;
    int N;
    while(scanf("%d",&n)!=EOF)
    {
        MAX=0;
        while(n--){
            scanf("%d",&a);
            if(MAX<prime[a]){
                MAX=prime[a];
                N=a;
            }
        }
        printf("%d\n",N);
    }
    return 0;
}




HDU 2710 Max Factor （筛选求素数）


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