Maximum GCD （stringstream）题解-白红宇

Maximum GCD （stringstream）题解

阅读量：6330 次

发布时间：2019-06-22

本文共 1194 字，大约阅读时间需要 3 分钟。

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

10 20 30 40

7 5 12

125 15 25

Sample Output

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
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           #include
            
              #include
             
              #include
              
               #include
               
                #include
                
                 #include
                 
                  #define INF 0x3f3f3f3f#define ll long longconst 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