Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28169 Accepted Submission(s): 10613
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
题意:
给出 T,代表有 T 组数据,每组数据开头给出数的个数 N,后给出这 N 个数。求出这些数的最小公因数。
思路:
数学。两个两个地求出最大公因数即可,最小公因数 = (a X b)/ 最大公约数。
AC:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int gcd (int a, int b) { return !(a % b) ? b : gcd(b,a % b); } int lcm (int a, int b) { return a / gcd (a,b) * b; } int main () { int t; scanf("%d", &t); while (t--) { int n, a, res = 1; scanf("%d", &n); while(n--) { scanf("%d", &a); res = lcm(res, a); } printf("%d\n", res); } return 0; }
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