hdu 1788(多个数的最小公倍数)

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hdu 1788(多个数的最小公倍数)

Chinese remainder theorem again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2415 Accepted Submission(s): 997 Problem Description 我知道部分同学最近在看中国剩余定理,就这个定理本身,还是比较简单的:假设m1,m2,…,mk两两互素,则下面同余方程…

hdu 1019 n个数的最小公倍数

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 f…

HDU 1019 （多个数的最小公倍数）

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1019 Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 61592 Accepted Submission(s): 23486 Problem Description The least comm…

n个数的最小公倍数

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 #include<stdio.h> int GCD(int num, int x) { if(num%x==0) retu…

HDOJ-ACM1019(JAVA) 多个数的最小公倍数

题意:求多个数的最小公倍数很简单,但是我一开始的做法,估计会让结果越界(超过int的最大值) import java.util.*; import java.io.*; public class Main{ public static void main(String[] arg){ Scanner scan = new Scanner(new BufferedInputStream(System.in)); int n =scan.nextInt(); int[] nums = new in…

HDU_2028——求多个数的最小公倍数

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 #include <cstdio> int gys(int a,int b) { ) return a; else r…

HDU - 1019-Least Common Multiple(求最小公倍数（gcd）)

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 f…