Lowest Common Multiple Plus(GCD+GBD)
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Lowest Common Multiple Plus
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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Problem Description
求n个数的最小公倍数。
Input
输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数。
Output
为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。
Sample Input
2 4 6
3 2 5 7
Sample Output
12
70
Author
lcy
Source
C语言程序设计练习(五)
一遍递推,并没有什么套路
#include<cstdio>#include<iostream>using namespace std;int gcd(int a, int b){ int i; while (a%b) { i = a%b; a = b; b = i; } return b;}int gbd(int a, int b)//求最小公倍数{ return a / gcd(a, b)*b;}int main(){ int n, i, s, a; while (~scanf("%d", &n)&&n) { scanf("%d", &s); for (i = 1; i<n; i++)//一路递推过去 { scanf("%d", &a); s = gbd(s, a); } printf("%d\n", s); } return 0;}
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