杭电 OJ 1019 多个数的最小公倍数

来源：互联网发布：80端口system占用编辑：程序博客网时间：2024/06/03 22:42

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

要求多个数的最小公倍数，先求多个数的最大公约数。

要求多个数的最大公约数，先求两个数的最大公约数，然后结果和第三个熟求最大公约数，以此类推。。

之后A,B,...,Z的最小公倍数=A*B*.....Z/(A,B,.....Z的最大公约数)即可。

本题A*B会超过int 范围，可以选择更大的数据类型，或者采用A/GCD(A,B)*B的写法。

还要考虑只有一个数的情况！就因为这个报超时，一直卡了好久

代码：

#include <iostream>#include <stdio.h>using namespace std;int LCM(int A,int B){int A1=A;int B1=B;int temp;while(B>0){temp=A%B;A=B;B=temp;}return A1/A *B1;}int main(){int n;int temp;int i=0;int c=0;int m;cin>>n;while(n--){scanf("%d",&m);if(m<=0){continue;}if(m==1){scanf("%d",&c);cout<<c<<endl;continue;}else{scanf("%d",&c);scanf("%d",&i);c=LCM(c,i);m--;m--;while(m--){cin>>i;c=LCM(i,c);}printf("%d\n",c);}}}

0 0