九度OJ 1056--最大公约数 1439--Least Common Multiple 【辗转相除法】

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概述

题目地址：http://ac.jobdu.com/problem.php?pid=1056

题目描述：

输入两个正整数，求其最大公约数。

输入：

测试数据有多组，每组输入两个正整数。

输出：

对于每组输入,请输出其最大公约数。

样例输入：

49 14

样例输出：

来源：

2011年哈尔滨工业大学计算机研究生机试真题

#include <stdio.h>
int gcd1 (int a, int b){
if (b == 0)
return a;
else
return gcd1 (b, a % b);
}
int gcd2 (int a, int b){
int tmp;
while (b != 0){
tmp = a;
a = b;
b = tmp % b;
}
return a;
}
int main (void){
int a, b;
while (scanf ("%d%d", &a, &b) != EOF){
printf ("%dn", gcd1 (a, b));
printf ("%dn", gcd2 (a, b));
}
return 0;
}

题目地址：http://ac.jobdu.com/problem.php?pid=1439

题目描述：

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

输入：

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

输出：

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

样例输入：

2
3 5 7 15
6 4 10296 936 1287 792 1

样例输出：

105
10296

#include <stdio.h>
int Gcd (int a, int b){
int tmp;
while (b != 0){
tmp = a;
a = b;
b = tmp % b;
}
return a;
}
int main(void){
int n;
int a;
int b;
int m;
while (scanf ("%d", &n) != EOF){
while (n-- != 0){
scanf ("%d%d", &m, &a);
--m;
while (m != 0){
scanf ("%d", &b);
a = a / Gcd (a, b) * b;
--m;
}
printf ("%dn", a);
}
}
return 0;
}

参考资料：维基百科 -- 辗转相除法