Lowest Common Multiple Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47977 Accepted Submission(s): 19890
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 6 3 2 5 7
Sample Output
12 70
Author
lcy
Source
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#include<stdio.h>
#include<String.h>
long divisor(long a,long b){
if(a%b==0){
return b;
}else{
return divisor(b,a%b);
}
}
int main(){
int n;
long a,b; //注意需要用长整形 否则a*b会越界
while(scanf("%d",&n) !=EOF&&n>0){
scanf("%d",&a);
if(n==1) {printf("%d\n",a);continue;}
for(int i =1;i<n;i++){
scanf("%d",&b);
a = a*b/divisor(a,b);
}
printf("%d\n",a);
}
return 0;
}