HDU-2028 Lowest Common Multiple Plus
来源:互联网 发布:淘宝达人账号怎么注册 编辑:程序博客网 时间:2024/05/15 23:54
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)
#include<stdio.h>int gcd(int a,int b){if(a%b==0) return b;else return gcd(b,a%b);}int lcm(int a,int b){return a/gcd(a,b)*b;}int a[10010];int main(){int n;while(scanf("%d",&n)!=EOF){int i,k=1;for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++){k=lcm(k,a[i]);}printf("%d\n",k);}return 0;}
0 0
- HDU 2028 Lowest Common Multiple Plus
- hdu - 2028 - Lowest Common Multiple Plus
- hdu 2028 Lowest Common Multiple Plus
- HDU 2028 - Lowest Common Multiple Plus (最小公倍数)
- [hdu 2028] Lowest Common Multiple Plus
- HDU 2028 Lowest Common Multiple Plus
- HDU 2028 Lowest Common Multiple Plus
- hdu 2028 Lowest Common Multiple Plus
- HDU 2028 Lowest Common Multiple Plus
- hdu 2028 Lowest Common Multiple Plus
- HDU - 2028 Lowest Common Multiple Plus
- hdu 2028 Lowest Common Multiple Plus
- HDU 2028 Lowest Common Multiple Plus
- hdu 2028 Lowest Common Multiple Plus
- HDU 2028 Lowest Common Multiple Plus
- HDU--2028-Lowest Common Multiple Plus
- HDU-2028 Lowest Common Multiple Plus
- HDU-2028 Lowest Common Multiple Plus
- 设计模式之策略
- 区块链技术相关理论
- 构建高性能web之路------mysql读写分离实战
- 百练 4116 拯救行动
- reuqest接收参数的几种方式
- HDU-2028 Lowest Common Multiple Plus
- java中的匿名内部类总结
- HDU1588 Gauss Fibonacci (矩阵快速幂+等比数列二分求和)
- C语言:指针及其运算
- 现代统计学——读书笔记
- jsp学习(二)
- Codeforces - Educational Codeforces Round 14C - Exponential notation(模拟)
- Spring4 MVC Hibernate4集成
- dbgrideh使用集合