UVa 11827 Maximum gcd:gcd& reading skills

11827-maximum GCD

Time limit:1.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=2927

Given the N integers, you have to find the maximum GCD (greatest common divisor) of every possible pair.

Input

The the "a" of input is a 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 this you have to find the maximum of GCD.

Output

For the maximum GCD of every possible pair.

Sample Input	Output for Sample Input
3 10 20 30 40 7 5 12 125 15 25	20 1 25

C Style:

/*0.012s*/
    
#include <cstdio>  
#include <algorithm>  
using namespace std;  
    
int num[100], n;  
    
int gcd (int a, int b)  
{return  
    b gcd (b, a% B): A;  
}  
    
int cal ()  
{  
    int i, j, maxn = 0;  
    for (i = 0; i < n-1. ++i) for  
        (j = i + 1; j < n; ++j)  
            MAXN = max (MAXN, gcd (num[i), num[j));  
    return MAXN;  
}  
    
int main ()  
{  
    int t;  
    char ch;  
    scanf ("%d\n", &t);  
    while (t--)  
    {  
        n = 0;  
        while (true)  
        {  
            scanf ("%d", &num[n++]);  
            while (ch = getchar ()) = = ")  
                ;  
            UNGETC (CH, stdin);  
            if (ch = = | | | ch = = 1) break;  
        }  
        printf ("%d\n", Cal ());  
    }  
    return 0;  
}

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C + + Style:

/*0.009s*/
    
#include <cstdio>  
#include <iostream>  
#include <sstream>  
#include <string>  
using namespace std;  
    
int num[100], n;  
string S;  
    
int gcd (int a, int b)  
{return  
    b gcd (b, a% B): A;  
}  
    
int cal ()  
{  
    int i, j, maxn = 0;  
    for (i = 0; i < n-1. ++i) for  
        (j = i + 1; j < n; ++j)  
            MAXN = max (MAXN, gcd (num[i), num[j));  
    return MAXN;  
}  
    
int main ()  
{  
    int t;  
    scanf ("%d\n", &t);  
    while (t--)  
    {  
        getline (CIN, s);  
        StringStream SS (s);  
        n = 0;  
        while (SS >> Num[n])  
            ++n;  
        printf ("%d\n", Cal ());  
    }  
    return 0;  
}

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