HDU-ACM課堂作業 Least Common Multiple
阿新 • • 發佈:2018-09-22
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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.<br>
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.<br>
Least Common Multiple
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 178 Accepted Submission(s) : 38
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.<br><br>
Sample Input 2 3 5 7 15 6 4 10296 936 1287 792 1
Sample Output 105 10296
Source East Central North America 2003, Practice 這裏INT-32類型的變量直接用int就可以啦
#include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> #include <iomanip> #include<math.h> using namespace std; int GCD(int num1, int num2) { if (num1%num2 == 0) return num2; else return GCD(num2, num1%num2); } int LCM(int a, int b) { int temp_lcm; temp_lcm = a / GCD(a, b)*b; //先除後乘,否則數據溢出導致WA return temp_lcm; } int main() { //數據組數 int n = 0; while (cin >> n) { while (n--) { //變量數 int k = 0; cin >> k; //將變量初始化 int b = 0; int lcm, a = 1; for (int i = 0; i < k; i++) { //將每次輸入的變量設置為b,計算a與b的LCM,隨後將LCM賦值給a cin >> b; lcm = LCM(a, b); a = lcm; } cout << lcm << endl; } } }
HDU-ACM課堂作業 Least Common Multiple