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Least Common Multiple

时间:2015-01-11 12:29:43      阅读:137      评论:0      收藏:0      [点我收藏+]

标签:printf   数据   测试   

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33455 Accepted Submission(s): 12636


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
105 10296
源代码:
#include <stdio.h> #include <stdlib.h> int grd(int u,int v) {   int r;   r=u%v;   if(r==0)     return v;   else     return grd(v,r);     } int main() {   int i,n,m,u,v,l;   scanf("%d",&n);   while(n--)   {     scanf("%d%d",&m,&u);     for(i=0;i<m-1;i++)     {       scanf("%d",&v);        l=grd(u,v);        u=u/l*v;//换一下,先除后乘,免得数据溢出                     }      printf("%d\n",u);              }   system("pause");    return 0;    }

Least Common Multiple

标签:printf   数据   测试   

原文地址:http://blog.csdn.net/zchlww/article/details/42609379

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