HDU——1019Least Common Multiple(多个数的最小公倍数)
Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42735 Accepted Submission(s): 16055
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
这题百度了下有出现n=1的情况,按我之前先取两个数得到第一个公倍数的做法会超时,n=1根本没无法输出。因此要重新写,顺便复习下gcd公式
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | #include<cstdio> #include<iostream> #include<algorithm> using namespace std; long long gcd( long long a, long long b) { return b?gcd(b,a%b):a; //还是记这个吧,简单易用 } int main() { int t; cin>>t; while (t--) { int n; long long a,tlcm,beg=1; //让beg初始化为1不影响结果并成为第0个数,这样一开始也就可以一个一个地求gcd scanf ( "%d" ,&n); for ( int i=0; i<n; i++) { scanf ( "%lld" ,&a); beg=(beg*a)/gcd(a,beg); } cout<<beg<<endl; } return 0; } |
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