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
就是求最小公倍数
代码:
#include<stdio.h> int a[10000]; int MAX(int n) { int i,max=0; for(i=0;i<n;i++) if(max<a[i]) max=a[i]; return max; } int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); int i,j,s=1,f=0; for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=2;i<=MAX(n);i++) { // printf("max=%d\n",MAX(n)); for(j=0;j<n;j++) if(a[j]%i==0) {a[j]/=i;f=1;} if(f==1) {s*=i;i=1;} f=0; } printf("%d\n",s); } return 0; }