最大公约数和最小公倍数

最新推荐文章于 2024-03-04 22:42:20 发布

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原文链接：http://www.cnblogs.com/zou20134585/p/gongyuegongbeishu.html

本文介绍了一个程序设计问题，该程序用于计算两个正整数的最小公倍数(LCM)和最大公约数(GCD)，适用于六年级学生的数学学习。输入包括多组数据，每组数据包含两个正整数，输出则对应每组数据的LCM和GCD。

Sixth Grade Math

Time Limit: 1000MS	Memory Limit: 65535KB
Submissions: 162	Accepted: 120

Description

In sixth grade, students are presented with different ways to calculate the Least Common Multiple (LCM) and the Greatest Common Factor (GCF) of two integers. The LCM of two integers a and b is the smallest positive integer that is a multiple of both a and b. The GCF of two non-zero integers a and b is the largest positive integer that divides both a and b without remainder.

For this problem you will write a program that determines both the LCM and GCF for positive integers.

Input

The first line of input contains a single integer N, (1 < = N < = 1000) which is the number of data sets that follow. Each data set consists of a single line of input containing two positive integers, a and b, (1< = a,b < = 1000) separated by a space.

Output

For each data set, you should generate one line of output with the following values: The data set number as a decimal integer (start counting at one), a space, the LCM, a space, and the GCF.

Sample Input

Sample Output

1 10 5 
2 161 1 
3 168 14

#include<stdio.h>
int main()
{
int n,i;
int a[1000],b[1000];
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d%d",&a[i],&b[i]);
}
for(i=1;i<=n;i++)
{
int da,xiao,yu;
if(a<b)
{
int t=a[i];
a[i]=b[i];
b[i]=t;
}
da=a[i];
xiao=b[i];
while(b[i])
{
yu=a[i]%b[i];
a[i]=b[i];
b[i]=yu;
}
printf("%d %d %d\n",i,da*xiao/a[i],a[i]);
}
return 0;
}

转载于:https://www.cnblogs.com/zou20134585/p/gongyuegongbeishu.html