UVa 11827 Maximum GCD：gcd&读入技巧

UVa 11827 Maximum GCD：gcd&读入技巧2014-07-24

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

C风格：

/*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
", &t）;while （t--）{n = 0;while （true）{scanf（"%d", &num[n++]）;while （（ch = getchar（）） == " "）;ungetc（ch, stdin）;if （ch == 10 || ch == -1） break;}printf（"%d
", cal（））;}return 0;}

C++风格：

/*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
", &t）;while （t--）{getline（cin, s）;stringstream ss（s）;n = 0;while （ss >> num[n]）++n;printf（"%d
", cal（））;}return 0;}

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