杭电acm--1019
2015-10-12 21:28
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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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
#include<stdio.h> #include<stdlib.h> #include<iostream> #include<cmath> #include<iomanip> //#define P 3.141592653 using namespace std; int function(int a, int b) { int a1 = a, b1 = b; int max, min, temp; if (a1 > b1) { max = a1; min = b1; } else { max = b1; min = a1; } temp = min; while (temp != 0) { temp = max%min; max = min; min = temp; } return max; } void main() { int i,n,arr[1000]; int a, b; int t = 0; cin >> n; while (n--) { int k; cin >> k; for (int j = 1; j <= k;j++) { cin >> a; arr[j] = a; } b = 1; for (i = 1; i + 1 <= k; i++) { b = arr[i] / function(arr[i + 1], arr[i])*arr[i + 1]; arr[i + 1] = b; } cout << arr[k] << endl; } system("pause"); }
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