（HDU1087）点击这里ProblemD：HumbleNumbers（HDU1058）点击这里ProblemE：MonkeyandBanana（HDU1069）点击这里ProblemF：数

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):22763    AcceptedSubmission(s):9950ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i

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。

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

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 {

Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17407    Accepted Submission(s):

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

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

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

  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

#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;

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17407    AcceptedSubmission(s):7565ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i

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

                                                   HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):13158    AcceptedSubmission(s):5715Problem

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

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,

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

HumbleNumbersTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):10640    AcceptedSubmission(s):4622ProblemDescriptionAnumberwhoseonlyprimefactorsare2,3,5or7i

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

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;

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:    

#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=

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

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

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];  //