SGU 499:Greatest Greatest Common Divisor

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPractice SGU 499

Description

Andrew has just made a breakthrough in sociology: he realized how to predict whether two persons will be good friends

 or not. It turns out that each person has an innerfriendship number (a positive integer). And thequality of friendship 

between two persons is equal to the greatest common divisor of their friendship number. That means there areprime 

people (with a prime friendship number) who just can't find a good friend, andWait, this is irrelevant to this problem. 

You are given a list of friendship numbers for several people. Find the highest possible quality of friendship among all 

pairs of given people.

Input

The first line of the input file contains an integer n () — the number of people to process. The next

  n lines contain one integer each, between 1 and (inclusive), the friendship numbers of the given people. 

All given friendship numbers are distinct.

Output

Output one integer — the highest possible quality of friendship. In other words, output the greatest greatest common 

divisor among all pairs of given friendship numbers.

Sample Input

sample input

sample output

49152516

5

//题意：找出n个数中某两个数的公约数的值最大。

//分析：直接暴力枚举时间复杂度O(n^2)是不允许的。题目中说明n个数的范围都在1000000以内，那么可以用一个数组

标记每个数据出现多少次。以最大的数据maxnum为上限开始枚举所有的数for(maxnum->1)的倍数，如果当前这个数

的倍数出现两次以上则得到最大公约数。整体时间复杂度约为10^6*(log(10^6)).

#include<stdio.h>#include<string.h>using namespace std;#define maxn 1000005#define max(p,q) (p>q?p:q)int num[maxn],a[maxn];int main(){int n,i,j,k;while(~scanf("%d",&n)){int maxp=1,ans;memset(num,0,sizeof(num));for(i=1;i<=n;i++){scanf("%d",&a[i]);maxp=max(maxp,a[i]);num[a[i]]++;}int flag=0;for(i=maxp;i>=1;i--){k=0;for(j=i;j<=maxp;j+=i){if(num[j])k+=num[j];if(k>1){printf("%d\n",i);flag=1;break;}}if(flag)break;}}return 0;}