HDU 1019 (多个数的最小公倍数)
Posted 若有恒,何必三更起五更眠;最无益,莫过一日曝十日寒。
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDU 1019 (多个数的最小公倍数)相关的知识,希望对你有一定的参考价值。
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1019
Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61592 Accepted Submission(s): 23486
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
Source
Recommend
题目意思:
求n个数的最小公倍数
直接暴力即可:暴力规则:求第一个和第二个元素的最小公倍数,然后拿求得的最小公倍数和第三个元素求最小公倍数,继续下去,直到没有元素
小技巧:
通过最大公约数求最小公倍数的时候,先除再乘,避免溢出
code:
#include<bits/stdc++.h> using namespace std; typedef long long LL; int gcd(int a,int b) { if (b==0) return a; return gcd(b, a%b); } int main() { //公式:a,b的最小公倍数等于a,b的乘积除以a,b的最大公约数 //直接暴力即可 int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); int a=1,cnt=1; int x; for(int i=0;i<n;i++) { scanf("%d",&x); cnt=a/gcd(a,x)*x;//先除后乘,避免溢出 a=cnt; } printf("%d\n",a); } return 0; }
以上是关于HDU 1019 (多个数的最小公倍数)的主要内容,如果未能解决你的问题,请参考以下文章
HDU 1019 Least Common Multiple