HDU-2028 Lowest Common Multiple Plus
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2016-07-06
HDU-2028 Lowest Common Multiple Plus
题目大意:求n个数的最小公倍数
解题思路:先求前两个数的最小公倍数,将前两数的最小公倍数与后面的数一一遍历求得最小公倍数。
#include <iostream>using namespace std;int main() { int n; int num[100]; int temp = 1; int lcm; int end; while ( cin >> n ) { for (int i = 0; i < n; i++) { cin >> num[i]; } while(1) { if(temp % num[1] == 0 && temp % num[0] == 0)//先求前两个数的最小公倍数 break; else temp++; } lcm = temp; for (int i = 2; i < n; i++) {//两两求最小公倍数 while(1) { if(temp % num[i] == 0 && temp % lcm == 0) break; else temp++; } lcm = temp; } cout << lcm << endl; lcm = 1; temp = 1; } return 0;}
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