最大公约数和最小公倍数
2013-11-17 21:34
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Sixth Grade Math
Time Limit: 1000MS | Memory Limit: 65535KB |
Submissions: 162 | Accepted: 120 |
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
3 5 10 7 23 42 56
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;
}
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