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  • Maximum GCD(UVA 11827)

    Problem: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

    题解:读入的时候处理一下,可以直接读入一个字符串,然后把数再按十进制还原存到数组中,或者直接用ungetc来退回一下。

    #include <bits/stdc++.h>
    using namespace std;
    int a[150];
    int main()
    {
        int n, maxx = -1;
        char op;
        while(~scanf("%d",&n))
        {
            while(n --)
            {
                maxx = -1;
                int i = 0;
                while(1)
                {
                    scanf("%d",&a[i++]);
                    while((op=getchar())==' '); // 如果是空格的话用ungetc退格
                    ungetc(op, stdin);
                    if(op == '
    ') break;
                }
                for(int j = 0; j < i; j ++)
                {
                    for(int k = j + 1; k < i; k ++)
                    {
                        if(__gcd(a[j],a[k]) > maxx) maxx = __gcd(a[j],a[k]);
                    }
                }
                printf("%d
    ",maxx);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lcchy/p/10139573.html
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