杭电acm1019题
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
#include <iostream>
using namespace std;
int zu(int a, int b) {
int r = 1;
while (r != 0) {
r = a%b;
a = b;
b = r;
}
return a;
}
int main() {
int n;
cin >> n;
int k;
int a, b;
while (n--) {
cin >> k;
cin >> a ;
for (int i = 1; i < k;i++) {
cin >> b;
a =(a/zu(a, b))*b;
}
cout << a << endl;
}
system("pause");
return 0;
}
#include <iostream>
using namespace std;
int zu(int a, int b) {
int r = 1;
while (r != 0) {
r = a%b;
a = b;
b = r;
}
return a;
}
int main() {
int n;
cin >> n;
int k;
int a, b;
while (n--) {
cin >> k;
cin >> a>>b ;
a== (a / zu(a, b))*b;
for (int i = 2; i < k;i++) {
cin >> b;
a =(a/zu(a, b))*b;
}
cout << a << endl;
}
system("pause");
return 0;
}
为什么第一个能ac，第二个不能ac
