杭电acm 1019 Least Common Multiple
2015-05-30 17:57
393 查看
Problem Description
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
Input
Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where
m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.
Output
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
Sample Input
Sample Output
//一道简单题错了这么多次,就是因为__int64最后数据太大的问题
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
Input
Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where
m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.
Output
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
Sample Input
2 3 5 7 15 6 4 10296 936 1287 792 1
Sample Output
105 10296
//一道简单题错了这么多次,就是因为__int64最后数据太大的问题
#include<cstdio> int fun(int max,int min) { int c,t; if(max<min) { t=min; min=max; max=t; } int a=max,b=min; while(min!=0) { c=max%min; max=min; min=c; } return max; } int main() { int t,n,i,a; scanf("%d",&t); while(t--) { scanf("%d",&n); __int64 res=1; for(i=1;i<=n;++i) { scanf("%d",&a); res=res*a/fun(res,a); } printf("%I64d\n",res); } return 0; }
相关文章推荐
- iOS 工程中创建 Prefix Header 文件 (xcode 6.0之后)
- JavaScript 使用穷举方式实现内容简繁转换
- XenServer 6.5实战系列之八:Creating a VM Template from an Existing VM
- Lucene基础(一)--入门
- 说文解字——傅里叶变换、拉普拉斯变换、Z变换 (上)
- ip地址与整数之间的转换
- android build系统
- java web中向PostgreSQL插入当前时间
- Android开发之handler(一)
- poj 3320 尺取法
- windows命令行的使用,去掉"半"字
- xiaoxia的vim配置
- C++用模板求解开方(你不得不知道的模板带给我们的运行效率)
- 2.5年, 从0->阿里
- 居里夫人
- Configuring HugePages for Oracle on Linux (x86-64)
- 手把手教你---进程资源分配
- linux 常用命令收集(持续更新)
- [leetcode] String to Integer
- Wampserver配置方法