Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53016 Accepted Submission(s): 20171
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
分析：多个数的最小公倍数，直接每次对两个进行lcm操作就好了，具体代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline ll gcd(ll a,ll b) 5 { 6 return b==0?a:gcd(b,a%b); 7 } 8 inline ll lcm(ll a,ll b) 9 { 10 return a*b/gcd(a,b); 11 } 12 ll a[100010]; 13 int main() 14 { 15 ll n; 16 while(scanf("%lld",&n)!=EOF) 17 { 18 while(n--) 19 { 20 ll m; 21 scanf("%lld",&m); 22 for(ll i=1;i<=m;i++) 23 scanf("%lld",&a[i]); 24 for(ll i=1;i<=m-1;i++) 25 a[i+1]=lcm(a[i],a[i+1]); 26 printf("%lld\n",a[m]); 27 } 28 } 29 return 0; 30 }
