hdu_1019Least Common Multiple(最小公倍数)
太简单了。。。题目都不想贴了
1 //算n个数的最小公倍数 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int gcd(int a, int b) 7 { 8 return b==0?a:gcd(b,a%b); 9 } 10 int lcm(int a, int b) 11 { 12 return a/gcd(a,b)*b; 13 } 14 int main() 15 { 16 int T; 17 scanf("%d",&T); 18 while(T--) 19 { 20 int n; 21 scanf("%d",&n); 22 int tm = 1; 23 int a; 24 for(int i = 0; i < n; i++){ 25 scanf("%d",&a); 26 tm = lcm(tm,a); 27 } 28 printf("%d\n",tm); 29 } 30 return 0; 31 }
 |
||||||||||
|
||||||||||
Least Common MultipleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45603 Accepted Submission(s): 17131 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
Recommend
|
||||||||||
|
然然
Mail 0
Control Panel 
Sign Out
