HDOJ 2710 Max Factor （筛素法求最大因子）

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5557    Accepted Submission(s): 1799

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

436384042

 

Sample Output

38

题意：给你n个数，求素因子最大的那个数，如果素因子大小相同，结果为最小的那个数

思路：在素数筛选过程中，用一个数组存储每个数的最大因子，因为因子为素数，所以说最后一个乘积所用的i是乘积数最大因子，然后过程比较就行了

ac代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define INF 0xfffffffusing namespace std;int v[1000100];int ans[1000100];void db(){memset(v,0,sizeof(v));v[1]=1;for(int i=2;i<1000100;i++){if(!v[i]){ans[i]=i;for(int j=i*2;j<1000100;j+=i)ans[j]=i,v[j]=1;}}}int main(){db();int n,a,i;while(scanf("%d",&n)!=EOF){int M=-INF;int Ans=-INF;for(i=0;i<n;i++){scanf("%d",&a);if(ans[a]>M){M=ans[a];Ans=a;}else if(ans[a]==M){if(a>Ans)Ans=a;}}printf("%d\n",Ans);}return 0;}