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V - Maximum GCDUVA - 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
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 );}}
