hdu1019

ccnu_2016_summer_week1(1)

pid=1019【hdu1019】ProblemDescriptionTheleastcommonmultiple(LCM)ofasetofpositiveintegersisthesmallestpositivei
Code_J_xer·2016-07-17 22:27

hdu1019——Least Common Multiple

LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):42668    AcceptedSubmission(s):16035ProblemDescriptionTheleastcommonmultiple(LCM)ofaset
qq_33110317·2016-02-03 20:00

ACM-简单题之Least Common Multiple——hdu1019

***************************************转载请注明出处： http://blog.csdn.net/lttree*************************************** Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)  &n
·2015-11-12 19:42

hdu1019 Least Common Multiple

View Code #include<iostream> #include<algorithm> using namespace std; #define max 100 int a[max]; int y(int a,int b) //求最大公约数 { if(b==0) return a; else return y(b,a%b)
·2015-11-11 01:50

HDU1019 Least Common Multiple

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1019 用辗转相除法求最小公倍数时要注意应该用a/y*b，若用a*b/y,则可能会数据溢出。 #include < iostream > #include  < vector > using   namespace  std;
·2015-10-31 08:54

多个数的最小公倍数 HDU1019

1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 5 using namespace std; 6 7 int gcd(int a,int b) 8 { 9 if(b==0) 10 return a; 11 e
·2015-10-28 07:38

HDU1019

LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):38901AcceptedSubmission(s):14699ProblemDescriptionTheleastcommonmultiple(LCM)ofasetofpositi
mrlry·2015-08-31 19:00

hdu1019 Least Common Multiple

LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):28021    AcceptedSubmission(s):10558ProblemDescriptionTheleastcommonmultiple(LCM)ofaset
svtter·2014-10-17 23:00

hdu1019 Least Common Multiple

LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):28021    AcceptedSubmission(s):10558ProblemDescriptionTheleastcommonmultiple(LCM)ofaset
svtter·2014-10-17 23:00

HDU1019 最小公倍数

LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):30295AcceptedSubmission(s):11460ProblemDescriptionTheleastcommonmultiple(LCM)ofasetofpositi
u013068502·2014-08-21 17:00

HDU1019 Least Common Multiple

#include#includeintgcd(inta,intb){  intr=a%b;  while(r)  {    a=b;b=r;r=a%b;  }  returnb;}intmain(){  intt;  scanf("%d",&t);  while(t--)  {    intn,i,ans;    longlongtemp=1;    scanf("%d",&n);    for(
AC_Gibson·2014-08-11 11:00

ACM-简单题之Least Common Multiple——hdu1019

***************************************转载请注明出处：http://blog.csdn.net/lttree***************************************LeastCommonMultipleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/
lx417147512·2014-05-28 13:00

HDU1019 Least Common Multiple

PS:如果开始求解前两个数字的最小公倍数，然后迭代求解下面的则TLE，如果开始求解第一个数字和1的LCM则0ms.#include #include #include #include usingnamespacestd; intgcd(inta,intb){ if(b==0)returna; elsereturngcd(b,a%b); } intmain() { intT,n,x; inta
wangwenhao00·2014-04-29 20:00

hdu1019 gcd和lcm

Least Common Multiple http://acm.hdu.edu.cn/showproblem.php?pid=1019 Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positive integer which i
richard_ma·2012-12-06 15:00

HDU1019 ( Least Common Multiple )

Problem:1019(LeastCommonMultiple)JudgeStatus:Accepted RunId:5548218Language:C++Author:ssun CodeRenderStatus:RenderedByHDOJC++CodeRenderVersion0.01Beta #include usingnamespacestd; longgcd(longm,longn
ssun125·2012-03-14 20:00

hdu1019

这题因为没看清题意，WA了7次妈的太恶心了！下面每行测试实例中，第一个数字竟然是后面数字的个数！之前一直把它也算进去，结果无限WA啊亏得还想各种办法弄“不指定个数的输入”问题，即用getchar()看是否到\n了，又用sstream串流类的还是WA,晕死不过还能收获个新用法，从数组中读数据，这玩意儿相当高档啊#include #include #include//串流类对象 intmain()
gneveek·2011-10-30 23:00

HDU 1019

HDU1019这是一道简单的数论知识题，即 a*b = gcd(a,b)*lcm(a, b);但是这里要注意一下如何求多个数的lcm 因为易在只有一个数的时候发生错误#include #include 
雪黛依梦·2010-09-06 20:00
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