Lowest Common Multiple Plus HDU2028
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Lowest Common Multiple Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59946 Accepted Submission(s): 24894
Problem Description
求n个数的最小公倍数。
Input
输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数。
Output
为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。
Sample Input
2 4 63 2 5 7
Sample Output
1270
long long a,b,t,n,x,c,d;
int gcd(int a,int b)
{
c=a;d=b;
while(a%b!=0)
{
t=a%b;
a=b;
b=t;
}
return c*d/b;
}
int main()
{
while(~scanf("%lld",&n))
{
t=1;
while(n--)
{
scanf("%lld",&x);
t=gcd(t,x);
}
printf("%lld\n",t);
}
}
#include<stdio.h>
int temp,n,i,t,a[50];
int gcd(int a,int b)
{
while(a%b!=0)
{
temp=a%b;
a=b;
b=temp;
}
return b;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
t=1;
for(i=0;i<n;i++)
{
t=(t*a[i])/gcd(t,a[i]);
}
printf("%d\n",t);
}
}
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