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 求最小公倍数算法: 最小公倍数=两整数的乘积÷最大公约数 __gcd(int,int)函数   返回值即是这两个数的最大公约数,使用时需要包含头文件#include<algorith…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34980    Accepted Submission(s): 14272 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.  …

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33474 Accepted Submission(s): 13683 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…

也称欧几里得算法 原理: gcd(a,b)=gcd(b,a mod b) 边界条件为 gcd(a,0)=a; 其中mod 为求余 故辗转相除法可简单的表示为: int gcd(int a, int b) { return b ==0? a:gcd( b, a% b); } 简洁而优雅. 例如:HDU 2028 Lowest Common Multiple Plus求n个数的最小公倍数. 最小公倍数=两数之积  /  最大公约数 这里防止中间过程溢出,先除以最大公约数,然后在求积. #includ…

解题思路:最近很忙,有点乱,感觉对不起自己的中国好队友.   好好调整,一切都不是问题,Just do it ! 代码: #include<cstdio> int gcd(int a, int b) { ) return a; return gcd(b, a % b); } int main() { int n, a, b, c; while(~scanf("%d", &n)) { scanf("%d", &a); ; i < n…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2028 思路:最一想到的就是暴力求解,从1开始一直到最后的答案,一直来除以给出的数列的数,直到余数为0:当然我们能升级,以输入的第一个数为初值,来除以数列,不行再进行翻倍操作,这样能减少循环的次数 开始用了break和continue ,最后发现复杂了. 复习哈C语言break continue break:强行退出循环,不再执行本次循环,即跳出此循环 (CPU直接结束这一个循环,运行下面的代码) con…

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 注意:用gcd求最小公倍数是,若两个数相乘后再除以最大公约数,可能会导致数据超限,从而wa,因此应改变一下运算的顺序 #include<…

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 // 数据溢出 #include<stdio.h> int cmd(int a, int b) { ; if(a<b) { t=a; a=b; b=t; } ) { t…

http://acm.hdu.edu.cn/showproblem.php?pid=2028 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   代码: #includ…

求一组数据的最小公倍数. 先求公约数在求公倍数.利用公倍数,连续求全部数的公倍数就能够了. #include <stdio.h> int GCD(int a, int b) { return b? GCD(b, a%b) : a; } inline int LCM(int a, int b) { return a / GCD(a, b) * b; } int main() { int T, m, a, b; scanf("%d", &T); while (T--)…

题目描述 正整数A和正整数B 的最小公倍数是指 能被A和B整除的最小的正整数值,设计一个算法,求输入A和B的最小公倍数. 输入描述:输入两个正整数A和B. 输出描述:输出A和B的最小公倍数. 输入例子: 5 7 输出例子: 35 思路:两个数的最小公倍数等于两个数的乘积除以最大公约数 最大公约数:分解质因数,找出其中相同的质因数,再将它们相乘,就得到了最大公约数,如果两数的质因数中,没有一个是相同的,那么它们的最大公约数就是1 ps 第二个方法太智障了  hold package huawei2…

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 <bits/stdc++.h> #define ll long long #define inf 100000000…

http://acm.hdu.edu.cn/showproblem.php?pid=2028 应该是比较简单的一道题啊...求输入的数的最小公倍数. 先用百度来的(老师教的已经不知道跑哪去了)辗转相除法求出两数的最大公因数,再将两数相乘并除以最大公因数即得到最小公倍数. 一开始我写的代码如下 #include<stdio.h> int gcd(int a, int b) { ) { return a; } return gcd(b, a%b); } int main() { int lop,…

链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2028 题目要求:就是求最大公倍数,我百度了一下,最好实现的算法就是:       公式法 由于两个数的乘积等于这两个数的最大公约数与最小公倍数的积.即(a,b)×[a,b]=a×b.所以,求两个数的最小公倍数,就可以先求出它们的最大公约数,然后用上述公式求出它们的最小公倍数. 例如,求[18,20],即得[18,20]=18×20÷(18,20)=18×20÷2=180.求几个自然数的最小公…

思考: 乘法爆咋数据.把int换成unsigned就过了,同时%d换成%u.求最大公约数和最小公倍数. #include<stdio.h> int gcd(unsigned x, unsigned y) { unsigned r; ) { x = y; y = r; } return y; } int lcm(unsigned x, unsigned y) { unsigned d; d = gcd(x, y); return (x*y / d); } int main() { int n;…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30907    Accepted Submission(s): 12528 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.  …

题目:Lowest common multiple plus 代码: #include<stdio.h> int common(int a,int b)//计算最大公约数 { int c=a%b,t=0; if(b>a) { t=b; b=a; a=t; } while(a%b!=0) { c=a%b; a=b; b=c; } return b; } int q[105]; int main() { int n,i,j,t=0; while(scanf("%d",&a…

原题链接在这里:https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/ 题目: Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3092 Least common multiple Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 Partychen like to do mathematical problems. One day, when he was doing on a least common m…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 64855 Accepted Submission(s): 24737 Problem Description The least common multiple (LCM) of a set of positive integers is the sm…

1143 Lowest Common Ancestor(30 分)The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. A binary search tree (BST) is recursively defined as a binary tree which has the following prop…

题目: Given the root and two nodes in a Binary Tree. Find the lowest common ancestor(LCA) of the two nodes. The lowest common ancestor is the node with largest depth which is the ancestor of both nodes. Example For the following binary tree: 4 / \ 3 7…

洛谷 SP14932 LCA - Lowest Common Ancestor 洛谷评测传送门 题目描述 A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia The lowest common ancesto…

Least Common Multiple (HDU - 1019) [简单数论][LCM][欧几里得辗转相除法] 标签: 入门讲座题解 数论 题目描述 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…

原题链接在这里:https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/ 题目: Given a rooted binary tree, return the lowest common ancestor of its deepest leaves. Recall that: The node of a binary tree is a leaf if and only if it has no childre…

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…

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w…

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has…

[题目] 输入二叉树中的两个结点,输出这两个结点在数中最低的共同父结点. 二叉树的结点定义如下:  C++ Code  123456   struct BinaryTreeNode {     int value;     BinaryTreeNode *left;     BinaryTreeNode *right; }; [分析] 求数中两个结点的最低共同结点是面试中经常出现的一个问题.这个问题有几个变种. [变种1] 第一个变种是二叉树是一种特殊的二叉树:查找二叉树.也就是树是排序过的,位…

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has…