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？……