hdu 1019 n个数的最小公倍数
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
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
题目的意思:有多组测试,每组有n个数,求这n个数的最小公倍数;
代码:
#include <iostream> using namespace std; long long n,m,i,j,k,a; int main() { cin>>n; while(n--){ cin>>m; cin>>k;m=m-1; while(m--) { cin>>a; j=k*a; while(k!=0) { i=a%k; a=k; k=i; } k=j/a; } cout<<k<<endl; } return 0; }思想是:是求两个数的最小公倍数延伸;比如现在有三个数,a,b,c;先求出a和b的最小公倍数d;然后再求d和a的最小公倍数e;e就是a,b,c的最小公倍数;
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