Maximum GCD （stringstream）题解

Given the N integers, you have to find the maximum GCD (greatest common divisor) of every possiblepair 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) positiveintegers 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

思路：

暴力gcd，之前小紫看到的stringstream可以拿出来用了，很方便啊，这里就用到了这个技巧

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack> 
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#include<sstream>
#define INF 0x3f3f3f3f
#define ll long long
const int N=16005;
const ll MOD=998244353;
using namespace std;
ll gcd(ll a,ll b){
	return b==0? a:gcd(b,a%b);
}
int main(){
	int n;
	ll num[110],x,MAX=-100000000;
	string s;
	scanf("%d",&n);
	getchar();
	while(n--){
		getline(cin,s); 
		int sum=0;
		MAX=-100000000;
		stringstream ss(s);
		while(ss>>x){
			num[sum++]=x;
		}
		for(int i=0;i<sum;i++){
			for(int j=i+1;j<sum;j++){
				MAX=max(MAX,gcd(num[i],num[j]));
			}
		}
		cout<<MAX<<endl;
	}
	return 0;
}

转载于:https://www.cnblogs.com/KirinSB/p/9409110.html