1662 - Least Common Multiple
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.
题目输入
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.
题目输出
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
输入/输出样例
题目输入
2 2 3 5 3 4 6 12
题目输出
15 12
C语言解答
#include<stdio.h> int gcd(int m,int n) { if(m%n) return gcd(n,m%n); return n; } int main() { int n,gbs,t,a; scanf("%d",&n); while(n--) { gbs=1; scanf("%d",&t); while(t--) { scanf("%d",&a); gbs=gbs*(a/gcd(gbs,a)); } printf("%d\n",gbs); } }
C++解答
#include<stdio.h> long long gcd(long long x,long long y) { if(!x||!y) return x>y?x:y; for(long long t;t=x%y;x=y,y=t); return y; } int main() { int n,m; long long a,b,g,l; scanf("%d",&n); while(n--) { scanf("%d%lld",&m,&a); if(m==1) { printf("%lld\n",a); continue; } scanf("%lld",&b); g=gcd(a,b); l=a*b/g; m-=2; while(m--) { scanf("%lld",&a); g=gcd(a,l); l=a*l/g; } printf("%lld\n",l); } return 0; }