hdu 1019 Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21158    Accepted Submission(s): 7899


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
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstring>
 6 using namespace std;
 7 int gcd(int a, int b){
 8   return b == 0 ? a : gcd(b, a%b);
 9 }
10 int main(void){
11   int t; 
12 #ifndef ONLINE_JUDGE
13   freopen("1019.in", "r", stdin);
14 #endif
15   scanf("%d", &t);{
16     while (t--){
17       int n; scanf("%d", &n);
18       int a, b, c; n--; scanf("%d", &a);
19       while (n--){
20         scanf("%d", &b);
21         a = b / gcd(a, b) * a;
22       }
23       printf("%d\n", a);
24     }
25   }
26   return 0;
27 }

这题和又和原来的不一样了……纠结,输入t的时候,不能用 while (~scanf("%d",&t)) ……唉,蛋疼……

又测试了一下,原来可以按照上面的用,发现如果改成long long int 就WA,为毛?……

posted on 2013-03-13 09:18  aries__liu  阅读(178)  评论(0编辑  收藏  举报