Wait,
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.
) — 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. sample input |
sample output |
4 9 15 25 16 |
5 |
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <cmath>
#include <queue>
#include <cstring>
#include <set>
#include <stack>
#include <string>
#define LL long long
#define maxn 100010
#define mod 1000000007
#define INF 2000000
#define MAX 16000010
#define eps 1e-6
using namespace std;
bool vi[maxn*10] ;
int Max ;
int max(int a ,int b){ return a > b ? a : b ;}
void insert( int n )
{
int m ,i ,k ;
m = sqrt(n+0.5) ;
for( i = 1 ; i <= m ;i++)if(n%i == 0)
{
k = n/i ;
if(vi[i])
{
Max = max(Max,i) ;
}
else vi[i] = true ;
if(vi[k]&&k != i)
{
Max = max(Max,k) ;
}
else vi[k] = true ;
}
}
int main()
{
int i ,m , tt , j , n ;
int L , R , mid ;
// freopen("in.txt","r",stdin) ;
// 枚举每个数的因子是可以过的
while( scanf("%d",&n) != EOF)
{
memset(vi,0,sizeof(vi)) ;
Max = 0 ;
for( i = 1 ; i <= n ;i++ ){
scanf("%d",&m) ;
insert(m) ;
}
printf("%d\n",Max) ;
}
return 0 ;
}
ssu 499 Greatest Greatest Common Divisor,布布扣,bubuko.com
ssu 499 Greatest Greatest Common Divisor
原文:http://www.cnblogs.com/20120125llcai/p/3596475.html