HDU 2028 Lowest Common Multiple Plus(水~)
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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
Solution
水题,两两求最小公倍数即可
Code
#include<cstdio>#include<iostream>using namespace std;int gcd(int x,int y){ if(x%y) return gcd(y,x%y); return y;}int main(){ int n; while(scanf("%d",&n)!=EOF) { int now,next; int flag=0; scanf("%d",&now); for(int i=1;i<n;i++) { scanf("%d",&next); now=next/gcd(now,next)*now; } printf("%d\n",now); } return 0;}
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