Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 46314    Accepted Submission(s): 17431 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   |   We have carefully selected several similar problems for you:   1005  1021  1008  1049  1108    #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<string> using namespace std; int gcd(int a,int b) { if(a<b)swap(a,b); return b?gcd(b,a%b):a; } int main() { int t,n;//t表示测试数组，n表示每组测试数据的个数 scanf("%d",&t); while(t--) { scanf("%d",&n); int num,Maxn;//num表示测试数据，Maxn表示最大公约数； scanf("%d",&num); Maxn=num; for(int i=1;i<n;i++) { scanf("%d",&num); Maxn=Maxn/gcd(Maxn,num)*num; } printf("%d\n",Maxn); } return 0; } 注: Maxn=Maxn/gcd(Maxn,num)*num; 的顺序如果是Maxn*num/gcd(Maxn,num)可能会因炸掉int而错误！！！