hdu1058

HDU动态规划题解目录

（HDU1087）点击这里ProblemD：HumbleNumbers（HDU1058）点击这里ProblemE：MonkeyandBanana（HDU1069）点击这里ProblemF：数
pxlsdz·2020-07-31 10:38

hdu1058 && hdu3199

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):22763    AcceptedSubmission(s):9950ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i
zyf_2014·2016-05-14 19:00

hdu1058 & hdu3199

hdu1058题意：一个数的因子只能是2,3,5,7，满足这样条件的数叫humble数，问第多少个是多少。
yexiaohhjk·2015-12-19 14:00

hdu1058 Humble Numbers

ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7iscalledahumblenumber.Thesequence1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,25,27,...showsthefirst20humblenumbers. Writeaprogramtofindandprin
wind_2008_06_29·2015-11-25 20:00

hdu1058(找第n个数,其因子只含2,3,5,7)

题目意思:找第n个数,其因子只含2,3,5,7. #include<stdio.h> int main() { int i,i1,i2,i3,i4,a[5850],min,min1; a[1]=1;i1=i2=i3=i4=1; for(i=2;i<=5842;i++)//打表算出拆分的最简因子只包含2,3,5,7四个数 {
·2015-11-13 19:09

HDU1058 Humble Numbers 【数论】

Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17407    Accepted Submission(s):
·2015-11-11 14:16

hdu1058丑数（优先队列、暴力打表）

hdu1058 题意：当一个数只有2、3、5、7这四种质因数时（也可以一种都没有或只有其中几种），这个数就是丑数，输出第 n 个丑数是多少； 其实并没有发现hdu把这道题放在 dp 专题里的意图，我的思路就是预处理出丑数数组
·2015-11-09 12:52

HDU1058——Humble Numbers

Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the firs
·2015-10-31 15:55

Humble Numbers（hdu1058）

Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14047    Accepted Submission(s): 6108 Problem D
·2015-10-27 14:47

DP题目列表，再续中。。。

  数塔  HDU2084数塔  最长公共子序 HDU1159 CommonSubsequence  最大子段和 HDU1003MaxSum  编辑距离 SDUT1225编辑距离  杭电的DP专题   HDU1058
wust_ZJX·2014-10-14 20:00

HDU1058 Humble Numbers

#include#includeusingnamespacestd;inthumble[6000];intMIN_4(inta,intb,intc,intd){  intmin1=a>b?b:a;  intmin2=c>d?d:c;  returnmin1>min2?min2:min1;}voidinit(){  humble[1]=1;  inta=1,b=1,c=1,d=1;  intans;
AC_Gibson·2014-09-15 21:00

HDU1058 Humble Numbers 【数论】

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17407    AcceptedSubmission(s):7565ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i
u012846486·2014-08-26 21:00

ACM-DP之Humble Numbers——hdu1058

HumbleNumbersProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7iscalledahumblenumber.Thesequence1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,25,27,...showsthefirst20humblenumbers.Writeaprogramt
lx417147512·2014-02-26 18:00

HDU1058 Humble Numbers

                                                   HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):13158    AcceptedSubmission(s):5715Problem
lsh670660992·2013-07-28 20:00

hdu1058

http://acm.hdu.edu.cn/submit.php?pid=1058思路想到了，但是不会处理～～～～～～～～看了别人的佩服额～#include #include #include #include usingnamespacestd; longlongnum[60000]; intb[4]={2,3,5,7}; longlongmin(longlonga,longlongb,long
JustSteps·2013-03-13 21:00

HDU1058 Humble Numbers

题目：HumbleNumbers humblenumber从1为"始祖",剩下的所有数,其实都是在此基础上乘以2,3,5,7演化出来的，代码主要语句:f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);#include #include usingnamespacestd; intf[5843],n; inti,j,k,l; intmin(inta,intb,
ACdreamers·2013-01-30 16:00

HDU1058：Humble Numbers

ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7iscalledahumblenumber.Thesequence1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,25,27,...showsthefirst20humblenumbers. Writeaprogramtofindandprin
libin56842·2013-01-24 21:00

hdu1058(dp)

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):10640    AcceptedSubmission(s):4622ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i
xiuyang_leiasp·2012-11-20 13:00

HDU1058

题目大意：输出第n个（素数因子只能为2、3、5、7(或者关系)）的合数思路：这个水题贴上来是因为第一次打超级长的表啊（哈哈，有意思），常规做法是用指数去四个for循环就搞定了 (打表程序)program：（由于数太大，我是分开来进行for循环的）#include#include#includeusingnamespacestd;intt[6000];intmain(){freopen("F.out
kg_second·2012-08-29 22:00

hdu1058 Humble Number

数据超级BT的一个题，WA了想不明白为什么错，看看别人的AC代码才知道，对于211这样的输出要用th,221要用st。那叫一个泪%>_ #include usingnamespacestd; intnum2[31]; intnum3[20]; intnum5[14]; intnum7[12]; intnum[6000]; intlen=0; intCalculate() { inti,j,k,w;
hss871838309·2012-03-08 20:00

hdu 1058(Humble Numbers)

hdu1058(HumbleNumbers) 1 /* 2 Author:    Leo.W 3 Descriptipn:寻找质因数仅为2、3、5、7其中若干的正整数     4 How to Do:    
冰王子Leo与ACM一起的成长·2012-03-07 14:00

hdu1058 优先队列与map判重、 补充手动堆排序做法、

#include #include #include #include usingnamespacestd; priority_queue,greater>q;//优先队列，值小的元素优先 mapmark;//map判重 intmain() { stringsign; longlongjs=1,re[5843],fx[4]={2,3,5,7};//re存储结果 re[1]=1; for(inti=
wl6965307·2012-02-09 20:00

hdu1058

杭电2010和2011级同学如何加入ACM集训队？HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):7113    AcceptedSubmission(s):3105ProblemDescriptionAnumberwhoseonly
fangzhiyang·2011-11-13 20:00

09-03 HDU_Steps3.2 简单DP HDU1003 HDU1159 HDU1087 HDU1160 HDU1058 HDU2084 HDU1176 HDU2571

HDUSTEPS3.2主要都是DP的入门题,最长XX序列,以及数塔问题3.2.1HDU1003MaxSum求连续区间使和最大首先将数列转化成前N项和的数列,这样[a,b]区间的和可以表示为sum[b]-sum[a-1]之后只要扫描一次数组,记录该位置之前的最小sum[]值,如果当前值减去该最小值得到的结果大于目前保存的最大值,则跟新最大值.#include usingnamespacestd; i
swm8023·2011-09-09 00:00

HDU 1058

HDU1058//重在理解方法：每次找到最小的 2 3 5  7 的因子数，之后利用所存数的下标的关系改变 #include using namespace std;int num[5843];  //
雪黛依梦·2010-09-11 20:00
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