HDOJ 1019 Least Common Multiple
2013-02-18 20:58
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Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20538 Accepted Submission(s): 7660
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
East Central North America 2003, Practice
Recommend
JGShining
求n个数的最小公倍数,就是先求前两个,用结果和第三个求最小公倍数。。。以此类推
#include <iostream> using namespace std; int gcd(int,int); int main(int argc, char *argv[]) { int a,n,m; long long s; cin>>m; for (int i=1;i<=m;++i){ cin>>n; s=1; for(int j=1;j<=n;++j){ cin>>a; s=s*a/gcd(s,a); } cout<<s<<endl; } return 0; } int gcd(int a,int b){ return b?gcd(b,a%b):a; }
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