HDU1019 Least Common Multiple 解题报告

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23605    Accepted Submission(s): 8824

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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296

 

Source

East Central North America 2003, Practice

 

Recommend

JGShining

 

#include<iostream>using namespace std;__int64 cmp(__int64 a,__int64 b){__int64 t;if(a>b) {t=a;a=b;b=t;}__int64 i;for(i=a;i>=1;i--){     if(a%i==0&&b%i==0) break;}return a*b/i;//两数相乘除以最大公约数位最小公倍数}int main(){__int64 r;//这题全部是64位scanf("%I64d",&r);while(r--){__int64 n;__int64 i;__int64 a[1001];scanf("%I64d",&n);for(i=0;i<n;i++){scanf ("%I64d",&a[i]);}for(i=0;i<n-1;i++){a[i+1]=cmp(a[i],a[i+1]);//很多数的最小公倍数求法}printf("%I64d\n",a[n-1]);}return 0;}