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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1019 解题思路:lcm(a,b)=a*b/gcd(a,b) 反思:最开始提交的时候WA,以为是溢出了,于是改成了long long,还是WA,于是就不明白了,于是就去看了discuss,发现应该这样来写 lcm(a,b)=a*gcd(a,b)*b;是为了以防a乘以b太大溢出,注意啊!!!!所以就先除再乘. #include<stdio.h> int gcd(int a,int b) { int 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…

主题 Calculate a + b 杭电OJ-1000 Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Mine #include <stdio.h> int main() { int a,b; while(~scanf("%d %d",&a,&b)) //多次…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1019 题目: Problem Description 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 an…

Problem Description 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 prob…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53016    Accepted Submission(s): 20171 Problem Description The least common multiple (LCM) of a set of positive integers is…

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…

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.  …

Problem Description 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 pr…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51959    Accepted Submission(s): 19706   Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positiv…

太简单了...题目都不想贴了 //算n个数的最小公倍数 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int gcd(int a, int b) { ?a:gcd(b,a%b); } int lcm(int a, int b) { return a/gcd(a,b)*b; } int main() { int T; scanf("%d",&…

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…

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…

Least Common Multiple Time Limit: 2 Seconds      Memory Limit: 65536 KB 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…

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. InputInput will consist of multiple problem instances. The fi…

题目链接:http://ac.jobdu.com/problem.php?pid=1439 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1439 Least Common Multiple.cpp // Jobdu // // Created by PengFei_Zheng on 10/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. /…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42735    Accepted Submission(s): 16055 Problem Description The least common multiple (LCM) of a set of positive integers is…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

***************************************转载请注明出处:http://blog.csdn.net/lttree*************************************** Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28975    …

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description 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. Fo…

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.  …

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): 25035    Accepted Submission(s): 9429 Problem Description The least common m…

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…

Robberies 点击打开链接 背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和- 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);  正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q[i].v)  当中,f[j]表示抢j块大洋的最大的逃脱概率,条件是f[j-q[i].money]可达,也就…

杭电acm题目分类版本1 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 1005 找规律(循环点) 1006 感觉有点BT的题,我到现在还没过 1007 经典问题,最近点对问题,用分治 1008 简单题 1009 贪心 1010 搜索题,剪枝很关键 1011 1012 简单题 1013 简单题(有个小陷阱) 1014 简单题 1015 可以看作搜索题吧 1016 经典的搜索 1017 简单数学题 1018 简单数学题 1019 简单数学题 1020 简单的字符串…

题目:Least common multiple 链接:http://acm.hdu.edu.cn/showproblem.php?pid=4913 题意:有一个集合s,包含x1,x2,...,xn,有xi=2^ai * 3^bi,然后给你a数组和b数组,求s所有子集合的最小公倍数之和.比如S={18,12,18},那么有{18},{12},{18},{18,12},{18,18},{12,18},{18,12,18},所以答案是174. 思路: 1. 最小公倍数,因为xi只包含两个质因子2.3…

抱着可能杭电的多校1比牛客的多校1更恐怖的想法 看到三道签到题 幸福的都快哭出来了好吗 1001  Maximum Multiple(hdoj 6298) 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6298 签到题 但是有考了一定的思维 清北大佬两分钟写出来真的让人望尘莫及啊…… 题意是给定一个n 可以由三个正整数相加得到 同时这三个正整数又是要被n可以整除 求这三个整数相乘的最大值 如果没有 则输出-1 既然题目没有要求三个正整数不能相等 则可以…

专注于C语言编程 C Programming Practice Problems (Programming Challenges) 杭电ACM题目分类 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093.1094.1095.1096.1097.1098.1106.1108.…

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…