hdu 5207(数论+枚举)

Greatest Greatest Common Divisor

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

ai,aj(i≠j)

 

Input

T, indicating the number of test cases. For each test cases, the first line contains an integer n, the size of the sequence. Next line contains n numbers, from a1 to an. 1≤T≤100,2≤n≤105,1≤ai≤105. The case for n≥104 is no more than 10.

 

Output

x: ans, x is the case number, starting from 1, ans

 

Sample Input

2 4 1 2 3 4 3 3 6 9

 

Sample Output

Case #1: 2 Case #2: 3

解题思路：cnt[i]记录因子为i的数的个数，接下来枚举i，找到最大的i使得cnt[i] >= 2即可。最开始用set超时了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 100005;
int n,cnt[maxn];

int main()
{
	int t,a,m,cas = 1;
	scanf("%d",&t);
	while(t--)
	{
		memset(cnt,0,sizeof(cnt));
		scanf("%d",&n);
		m = 0;
		for(int i = 1; i <= n; i++)
		{
			scanf("%d",&a);
			m = max(m,a);
			int k = sqrt(a + 0.5);
			for(int j = 1; j <= k; j++)
				if(a % j == 0)
				{
					cnt[j]++;
					cnt[a/j]++;
				}
			if(k * k == a)
				cnt[k]--;
		}
		for(int i = m; i >= 1; i--)
			if(cnt[i] >= 2)
			{
				printf("Case #%d: %d\n",cas++,i);
				break;
			}
	}
	return 0;
}