【杭电oj2028】Lowest Common Multiple Plus
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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): 52009 Accepted Submission(s): 21589
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
Author
lcy
Source
C语言程序设计练习(五)
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#include<stdio.h>#include<algorithm>using namespace std;int min(int a,int b);int main() { int n; while(scanf("%d",&n)==1) { int a[10000],k,ret; for(int l=0; l<n; l++) scanf("%d",&a[l]); sort(a,a+n); k=min(a[n-1],a[n-2]); for(int l=n-3; l>=0; l--) { ret=min(a[l],k); k=ret; } printf("%d\n",k); } return 0;}int min(int a,int b) { int i; for(int i=a; i<=a*b; i++) if(i%a==0&&i%b==0) return i;}
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