ACM--最大公约数--HDOJ 1019--Least Common Multiple--水 Least Common Multiple

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概述

HDOJ题目地址：传送门

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 45286 Accepted Submission(s): 17030

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

题意题解：求给定你的一排数的最大公约数，这里特别要注意的一点就是需要用__int64，否则会WA

#include<stdio.h>
#include<iostream>
using namespace std;
__int64 lcm(__int64 a,__int64 b){
__int64 temp=a*b;
__int64 r=0;
while(b){
r=a%b;
a=b;
b=r;
}
return temp/a;
}
int main(){
int i ,n;
cin>>n;
while(n--){
int t;
__int64 a;
int b;
cin>>t;
cin>>a;
for(int i=1;i<t;i++){
cin>>b;
a=lcm(a,b);
}
printf("%I64dn",a);
}
}