HDU(杭电OJ)

HDU 又一版 A+B

题目：输入两个不超过整型定义的非负10进制整数A和B(#include#include//http://acm.hdu.edu.cn/showproblem.php?
yunshouhu·2020-09-13 00:21

HDU 2186 取整小技巧

原题链接：http://acm.hdu.edu.cn/showproblem.php?
Special__Yang·2020-09-13 00:48

160个练手CrackMe-047

A3AF214000movdwordptrds:[0x4021AF],eax;|0040105C.6A00push0x0;|/hTemplateFile=NULL0040105E.686F214000pushDueList
Mnnk·2020-09-13 00:17

hdu1877 又一版 A+B

#include#include#includeusingnamespacestd;intmain(){intA,B,m;chars[1005];while(cin>>m,m){cin>>A>>B;memset(s,0,sizeof(s));itoa(A+B,s,m);cout<
Flynn_curry·2020-09-13 00:12

HDU4465(数学期望+大数处理)

CandyTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):507AcceptedSubmission(s):232SpecialJudgeProblemDescriptionLazyChildisalazychildwholikescandyverymuch.Desp
晓风残月xj·2020-09-12 23:02

hdu5534—Partial Tree（完全背包+思维，转化的超级巧妙哇~）

PartialTreeTimeLimit:2000/1000MS(Java/Others)MemoryLimit:262144/262144K(Java/Others)TotalSubmission(s):2179AcceptedSubmission(s):1086ProblemDescriptionInmathematics,andmorespecificallyingraphtheory,at
sdau_blue·2020-09-12 23:20

HDU 4433 locker（DP）（ UVA 1631 - Locker）

题目链接：点击打开链接题意：有一个n位密码锁，每一位都是0~9，可以循环旋转，每次可以让1~3个相邻数字同时往上或者往下旋转一格。输入初始状态和终止状态，问最少需要转几次。思路：很显然是个DP题目，由于每次可以让相邻的1~3个数字同时旋转，所以状态的表示上就要考虑相邻3个位置。那么可以用d[i][a][b][c]表示当前到了第i位，第i位上当前是a，i+1位是b，i+2位是c。那么显然必须让i位上
AC_Arthur·2020-09-12 23:01

HDU 4433 locker 第37届ACM/ICPC 天津赛区现场赛C题（DP）

lockerTimeLimit:6000/3000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):248AcceptedSubmission(s):87ProblemDescriptionApasswordlockerwithNdigits,eachdigitcanberotatedto0-9circul
weixin_33693070·2020-09-12 23:48

带限制求最小价值的完全背包 HDU1114

1#include2#include34usingnamespacestd;56intv[510];7intw[510];8intdp[10010];910intmain()11{12intt;13cin>>t;14while(t--)15{16inta,b;17cin>>a>>b;18intn;19cin>>n;20for(inti=0;i>v[i]>>w[i];23}24for(inti=1;
weixin_30237281·2020-09-12 23:56

hdu 5176 The Experience of Love(带权并查集+思维)

题意：给一棵树，求任意｛两点路径上的最大边权值-最小边权值｝的总和。解法：sigma(maxVal[i]−minVal[i])=sigma(maxVal)−sigma(minVal);所以我们分别求所有两点路径上的最大值的和，还有最小值的和。再相减就可以了。求最大值的和的方法用带权并查集，把边按权值从小到大排序，一条边一条边的算，当我们算第i条边的时候权值为wi,两点是ui,vi,前面加入的边权值
hadis_fukan·2020-09-12 23:58

hdu 5175 Misaki's Kiss again (抑或运算+公式变形)

题意：找出1-n之间所有的m使得gcd(n,m)=n^m。分析：令n^m=k，可以推出n^k=m，m^k=n。则由gcd(n,m)=n^m=k可以推出gcd(n,n^k)=k且k是n的约数。故找出n的所有约数，判断是否满足gcd(n,n^k)=k即可。n^k=0要舍去，因为此时k=n，不满足gcd(n,n)=(n^n)。而且抑或运算得到的数可能变大，如1^2=3，故要判断(n^k)#include
hadis_fukan·2020-09-12 23:57

hdu 4465 Candy （快速排列组合 ）

CandyTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):2115AcceptedSubmission(s):910SpecialJudgeProblemDescriptionLazyChildisalazychildwholikescandyverymuch.Des
1A_coder·2020-09-12 23:19

hdu 5534 Partial Tree（完全背包）

题目链接：hdu5534PartialTree解题思路首先度的总和为2(n-1)，并且每个节点度不为0。如果用二维dp[i][j]表示第i个节点还剩j个度的最优值，是没问题，但是复杂度为o(n3)。
JeraKrs·2020-09-12 23:43

HDU——1846 Brave Game（巴什博弈）

题目链接：#include#include#include#include#include#include#include#include#include#includeusingnamespacestd;#definelllonglongintmain(){intt,n,m;cin>>t;while(t--){cin>>n>>m;intmod=n%(m+1);if(mod>=1)cout<<"f
ZMST·2020-09-12 23:21

HDU——6015 Skip the Class

题目链接：一个sort排序但是要注意是否出现了大于等于三次所以需要map把string对应到int计次数#include#include#include#include#include#include#include#include#includeusingnamespacestd;#definelllonglongstructnode{chars[10];intval;}s[110];boolc
ZMST·2020-09-12 23:20

HDU4989——Summary（set）

题目链接：#include#include#include#include#include#include#include#include#include#includeusingnamespacestd;#definelllonglonglla[110];intmain(){intn;while(~scanf("%d",&n)){sets;s.clear();for(inti=0;i::iter
ZMST·2020-09-12 23:20

HDU 5534 Partial Tree 完全背包

Inmathematics,andmorespecificallyingraphtheory,atreeisanundirectedgraphinwhichanytwonodesareconnectedbyexactlyonepath.Inotherwords,anyconnectedgraphwithoutsimplecyclesisatree.Youfindapartialtreeonthew
紫芝·2020-09-12 23:30

hdu4465

CandyTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):2705AcceptedSubmission(s):1199SpecialJudgeProblemDescriptionLazyChildisalazychildwholikescandyverymuch.De
yawen_2016·2020-09-12 22:03

hdu4433 locker（dp）

hdu4433题目给出两个串，每次可以选择连续的1-3个数字，进行同时加1或者同时减1，问最少经过多少次操作，将一个串转变为另外一个串.思路dp[i][j][k]表示前i个已经完全匹配，而这时候，第i+
pibaixinghei·2020-09-12 22:17

Hdu 4433 locker【思维+Dp】

lockerTimeLimit:6000/3000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):2082AcceptedSubmission(s):949ProblemDescriptionApasswordlockerwithNdigits,eachdigitcanberotatedto0-9circ
mengxiang000000·2020-09-12 22:02

HDU 5534 - Partial Tree（完全背包）【真*好题】

PartialTreeTimeLimit:2000/1000MS(Java/Others)MemoryLimit:262144/262144K(Java/Others)TotalSubmission(s):2087AcceptedSubmission(s):1035ProblemDescriptionInmathematics,andmorespecificallyingraphtheory,at
辞树 LingTree·2020-09-12 22:57

HDU 4433 locker

ProblemDescriptionApasswordlockerwithNdigits,eachdigitcanberotatedto0-9circularly.Youcanrotate1-3consecutivedigitsupordowninonestep.Forexamples:567890->567901(byrotatingthelast3digitsup)000000->000900
_Occult_·2020-09-12 22:20

Candy----HDU4465----数学题

题目地址：http://acm.hdu.edu.cn/showproblem.php?
dr5459·2020-09-12 22:07

HDU4465 Candy

题意：有两个箱子，里面放有相同数量的糖果，每次从第一个箱子里拿的概率是p，从第二个箱子里拿的是1-p，问拿完一个箱子，另一个箱子剩下糖果数量的期望。第一眼看到这题以为是概率DP，然后推了下发现不是，就一个很好推的公式，但是那个组合数太大，无法求出。。然后就没有然后了。。公式是C(n+i,i)*(p^(n+1)*(1-p)^i+p^i*(1-p)^(n+1))*(n-i)。。n+1次方的原因是，先取
eeeaaaaa·2020-09-12 22:55

HDU4465-数学期望

链接：http://acm.hdu.edu.cn/showproblem.php?
d12155214552·2020-09-12 22:37

UVA 1631 Locker（HDU 4433）（DP）

ApasswordlockerwithNdigits,eachdigitcanberotatedto0-9circularly.Youcanrotate1-3consecutivedigitsupordowninonestep.Forexamples:567890→567901(byrotatingthelast3digitsup)000000→000900(byrotatingthe4thdig
JOY酷酷·2020-09-12 22:10

hdu_4465_Candy

LazyChildisalazychildwholikescandyverymuch.Despitebeingveryyoung,hehastwolargecandyboxes,eachcontainsncandiesinitially.Everydayhechoosesoneboxandopenit.Hechoosesthefirstboxwithprobabilitypandthesecond
awang50683·2020-09-12 22:37

最小生成树Prim模板

e.g.畅通工程再续http://acm.hdu.edu.cn/showproblem.php?pid=1875Freckleshttp://poj.org/problem?
Rachel-Zhang·2020-09-12 22:54

HDU 4465 快速全排列

#include#includeconstintmaxn=4E5+10;intn,kase;doublefun[maxn]={0},p;voidinit(){for(inti=1;i
Tczxw·2020-09-12 22:04

hdu 4465 Candy

CandyTimeLimit:2000/1000ms(Java/Other)MemoryLimit:32768/32768K(Java/Other)TotalSubmission(s):6AcceptedSubmission(s):5SpecialJudgeFont:TimesNewRoman|Verdana|GeorgiaFontSize:←→ProblemDescriptionLazyChil
Singular__point·2020-09-12 22:58

hdu 4465

计算组合数最大的困难在于数据的溢出，对于大于150的整数n求阶乘很容易超出double类型的范围，那么当C(n,m)中的n=200时，直接用组合公式计算基本就无望了。另外一个难点就是效率。对于第一个数据溢出的问题，可以这样解决。因为组合数公式为：C(n,m)=n!/(m!(n-m)!)为了避免直接计算n的阶乘，对公式两边取对数，于是得到：ln(C(n,m))=ln(n!)-ln(m!)-ln((n
SYLG_li·2020-09-12 22:25

HDU - 5534 Partial Tree (完全背包，物品价值包含负数）

Inmathematics,andmorespecificallyingraphtheory,atreeisanundirectedgraphinwhichanytwonodesareconnectedbyexactlyonepath.Inotherwords,anyconnectedgraphwithoutsimplecyclesisatree.Youfindapartialtreeonthew
vocaloid01·2020-09-12 21:33

hdu4465 Candy(快速排列组合+概率)

CandyTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):679AcceptedSubmission(s):295SpecialJudgeProblemDescriptionLazyChildisalazychildwholikescandyverymuch.Desp
爱情魔法师·2020-09-12 21:29

hdu4433 locker 密码锁（枚举）

lockerTimeLimit:6000/3000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):883AcceptedSubmission(s):374ProblemDescriptionApasswordlockerwithNdigits,eachdigitcanberotatedto0-9circu
爱情魔法师·2020-09-12 21:28

HDU 1219 AC Me

ACMeTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):7582AcceptedSubmission(s):3410ProblemDescriptionIgnatiusisdoinghishomeworknow.Theteachergiveshimsomearticl
wpfengqi·2020-09-12 21:21

HDU 1209 Clock

ClockTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):2754AcceptedSubmission(s):845ProblemDescriptionThereisananalogclockwithtwohands:anhourhandandaminutehand.
wpfengqi·2020-09-12 21:21

HDU 1000 A + B Problem

A+BProblemTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):240144AcceptedSubmission(s):70866ProblemDescriptionCalculateA+B.InputEachlinewillcontaintwointegersA
wpfengqi·2020-09-12 21:50

HDU 1205 吃糖果

吃糖果TimeLimit:6000/3000MS(Java/Others)MemoryLimit:65535/32768K(Java/Others)TotalSubmission(s):12695AcceptedSubmission(s):3670ProblemDescriptionHOHO，终于从Speakless手上赢走了所有的糖果，是Gardon吃糖果时有个特殊的癖好，就是不喜欢将一样的糖果
wpfengqi·2020-09-12 21:50

HDU 1202 The calculation of GPA

ThecalculationofGPATimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):12828AcceptedSubmission(s):2952ProblemDescription每学期的期末，大家都会忙于计算自己的平均成绩，这个成绩对于评奖学金是直接有关的。国外
wpfengqi·2020-09-12 21:50

HDU 1012 u Calculate e

uCalculateeTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):19373AcceptedSubmission(s):8458ProblemDescriptionAsimplemathematicalformulaforeiswherenisallowedtog
wpfengqi·2020-09-12 21:50

HDU 1001 Sum Problem

SumProblemTimeLimit:1000/500MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):170123AcceptedSubmission(s):40500ProblemDescriptionHey,welcometoHDOJ(HangzhouDianziUniversityOnlineJu
wpfengqi·2020-09-12 21:50

最长公共子序列LCS，最长递增子序列LIS

详情可看HDU1159。我们可以根据递推的方式得到最长的长度，如果想得到子序列也只需要记忆化一下每个点是从哪个点递推得到的即可。
weixin_34327223·2020-09-12 20:29

HDU4768：Flyer [ 二分的奇妙应用 好题 ]

传送门FlyerTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1718AcceptedSubmission(s):622ProblemDescriptionThenewsemesterbegins!Differentkindsofstudentsocietiesar
weixin_33772645·2020-09-12 20:36

hdu 4190(二分)

题意:给你N个城市每个城市有相应的人口，有B个选票箱,要求每个城市都至少有一个选票箱，且先要使N个城市中得选票箱的票数的最大值最小化。解题思路:二分结果，求上界注意:无#include#include#include#include#defineMAXN0500010#defineMAXN15000010#definer1(x)(x)>>1intN,B;inta[MAXN0];intsolve()
JoyGatsby·2020-09-12 20:57

HDU 4768 Flyer(二分)

FlyerTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1445AcceptedSubmission(s):510ProblemDescriptionThenewsemesterbegins!Differentkindsofstudentsocietiesareal
lab104_yifan·2020-09-12 20:35

HDU4768：Flyer(二分)

ProblemDescriptionThenewsemesterbegins!Differentkindsofstudentsocietiesarealltryingtoadvertisethemselves,bygivingflyerstothestudentsforintroducingthesociety.However,duetothefundshortage,theflyersofaso
键盘上的舞者·2020-09-12 19:13

Flyer(二分 HDU4768)

FlyerTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):2009AcceptedSubmission(s):736ProblemDescriptionThenewsemesterbegins!Differentkindsofstudentsocietiesareal
绝尘花遗落·2020-09-12 19:11

HDU1528（二分图匹配）

题意：一个人把手牌放在桌子上，另一个人的牌还在手中，他需要调整出牌顺序，是自己能赢得更多的局数，输出最大能赢得局数。说白了就是个二分图匹配，不过那两个点相连需要自己写函数判断。code：#include#include#include#include#include#include#include//a&3==a%4usingnamespacestd;constdoubleeps=1e-8;con
Cai_Haiq·2020-09-12 18:13

HDU 3215 The first place of 2^n （数论-水题）

Thefirstplaceof2^nProblemDescriptionLMYandYYaremathematicsandnumbertheorylovers.Theyliketofindandsolveinterestingmathematicproblemstogether.OnedayLMYcalculates2nonebyone,n=0,1,2,…andwritestheresultson
你狗·2020-09-12 17:02

iOS 之UITextFiled/UITextView小结

windkisshao/article/details/213985211.自定方法,用于移动视图-(void)moveInputBarWithKeyboardHeight:(float)_CGRectHeightwithDuration
weixin_30621959·2020-09-12 17:19
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