V - Maximum GCD UVA - 11827

V - Maximum GCD UVA - 11827

description

Given the N integers, you have to find the maximum GCD (greatest common divisor) of every possible
pair of these integers.
Input
The first line of input is an integer N (1 < N < 100) that determines the number of test cases.
The following N lines are the N test cases. Each test case contains M (1 < M < 100) positive
integers that you have to find the maximum of GCD.
Output
For each test case show the maximum GCD of every possible pair.
Sample Input
3
10 20 30 40
7 5 12
125 15 25
Sample Output
20
1
25

题意：给一组数求两个数之间的最大公约数。

题解：直接暴力求即可，需要用sstream流来进行数据的读入。

AC代码：

/*
* @Author: 王文宇
* @Date:   2017-10-28 15:30:11
* @Last Modified by:   王文宇
* @Last Modified time: 2017-10-28 16:50:34
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <sstream>
using namespace std;
const int maxn =107;
int a[maxn];
int t,n;
int b[maxn][maxn];
int gcd(int x,int y)
{
	if(y==0)return x;
	return gcd(y,x%y);
}
int main()
{
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		n=1;
		string s;
		getline(cin,s);
		stringstream ss(s);
		while(ss>>a[n])
			{
				++n;
			}
		int max1 = 0;
		for(int i=1;i<n;i++)
		{
			for(int j=i+1;j<n;j++)
			{
				max1 = max(max1,gcd(a[i],a[j]));
			}
		}
		printf("%d\n",max1 );
	}
}