杭电acm1019 hdu-acm-101解题报告

最新推荐文章于 2020-03-28 21:59:50 发布

yangjing1003004612

最新推荐文章于 2020-03-28 21:59:50 发布

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本文链接：https://blog.youkuaiyun.com/yangjing1003004612/article/details/38372003

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题目链接：

点击打开题目链接

原题：

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29996 Accepted Submission(s): 11345

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

   
   
    
    2
3 5 7 15
6 4 10296 936 1287 792 1

Sample Output

题目大意：

就是第一次输入一个数T，然后有T组测试数据；

然后输入，n，求n个数的公共倍数；

代码如下：

#include<stdio.h>
int ys(int a,int b)
{
	return !b?a:ys(b,a%b);
}
int main()
{
	int T,n,b,bs,i;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&bs);
		for(i=1;i<n;i++)
		{
			scanf("%d",&b);
			bs=bs/ys(bs,b)*b;//应该先除后乘，免得数据溢出，比如乘了以后超过了32位数； 
		}
		printf("%d\n",bs);
	}
	return 0;
	
}

运行结果：