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): 35196 Accepted Submission(s): 14377
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源代码:#include<stdio.h> int f(int i,int j){ int a,b,c; int tempi = i,tempj = j; if(i<j) { b=i; i=j; j=b; } while(j!=0) { c=i%j; i=j; j=c; } return tempi/i*tempj;}int main() { int n,i,t,a[100]; while(~scanf("%d",&n)) { if(n==0) continue; t=1; for(i=0;i<n;i++) { scanf("%d",&a[i]); t=f(a[i],t); } printf("%d\n",t); } return 0; }
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