joj

joj 1171

 水题，做1089太郁闷了所以刷了到水题，代码很笨，基本没有参考价值，只为看一眼快把我折磨死的气球。。。。。。。。。。。#include#includeusingnamespacestd;charstr1[1005],str2[1005];intmain(){   while(scanf("%s%s",str1,str2)!=EOF)   {        intlen1,len2,i=0;   
qiankun1993·2011-11-06 18:00

joj 2343

 我这辈子都会记住这个定理：n元循环群（a）中，元素a^m是（a）的生成元的充要条件是（n，m）=1.哎，郁闷。。。。#include#includeusingnamespacestd;intmain(){   intn,m;   while(scanf("%d%d",&n,&m)!=EOF)   {       inta,b;       if(n>m)       {          a=n
qiankun1993·2011-11-06 17:00

joj 1387

又见0-1背包，长度既是价值也是体积。#include#includeusingnamespacestd;intdp[100000];intcd[105];intmain(){   intcd_sto;   while(scanf("%d",&cd_sto)!=EOF)   {       intn;       scanf("%d",&n);       inti,j;       for(i=
qiankun1993·2011-11-04 20:00

joj 1078 hdu 1116

欧拉回路在加上并查集判断连通性，把每一个单词当做一个边,话说我一开始当做h路做的。。。。。#include#include#includeusingnamespacestd;intfather[26],rank[26],r[26],c[26],mark[26];intfind(intx){  //coutrank[y])    {        father[y]=x;    }    else 
qiankun1993·2011-11-03 23:00

joj 1317

 各种序。。。。。。。。。。。#include#include#includeusingnamespacestd;voidtree(char*s1,char*s2){    if(s1[0]=='\0')        return;    elseif(s1[1]=='\0')    {        printf("%c",s1[0]);        return;    }    inti
qiankun1993·2011-10-30 15:00

joj2035

一开始超时，怎么改都超时，后来果断把stl换成纯数组，竟然AC了，莫非stl的操作很费时间吗，不解：之前的超时stl版：#include#include#includeusingnamespacestd;intmark[10005];intmain(){   intn;      while(scanf("%d",&n),n>0)   {     if(n==0)       continue; 
qiankun1993·2011-10-30 10:00

joj1968

 1968:CommonSubsequenceResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE1s4096K761222StandardAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anotherseq
hechenghai·2011-10-29 00:00

joj 1329 二叉树

给出中根序列和后根序列求路径值最小的叶节点，各种队，各种栈（话说一开始理解错题了，英语不好的悲哀） #include#include#includeusingnamespacestd;intmmaaxx;intk;voidtree(stacks1,stacks2,intm){    if(s1.size()==1&&s2.size()==1)    {       if((m+s1.top())p
qiankun1993·2011-10-28 21:00

joj1829

 1829:CandiesResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K922220StandardAllchildrenlikecandies.OnedayagroupofchildrenaredividingMcandiesbythefollowingrules.1.Thecandiesarenumberedwith1,2,3…M.2
hechenghai·2011-10-27 23:00

joj2529

 2529:ChorusResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE1s65536K551154StandardN位同学站成一排，音乐老师要请其中的(N-K)位同学出列，使得剩下的K位同学排成合唱队形。    合唱队形是指这样的一种队形：设K位同学从左到右依次编号为1，2…，K，他们的身高分别为T1，T2，…，TK，  则他们的身高满足T1…>TK(
hechenghai·2011-10-26 11:00

JOJ1043：Atlantis 离散化+扫描线

JOJ1043：Atlantis离散化+扫描线 1043:AtlantisResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K431155StandardThereareseveralancientGreektextsthatcontaindescriptionsofthefabledislandAtlantis.Someofthesetext
第七天堂·2011-10-25 23:00

JOJ1040：Trees（卡特兰数+递归）

JOJ1040：Trees（卡特兰数+递归）Wecannumberbinarytreesusingthefollowingscheme:Theemptytreeisnumbered0.Thesingle-nodetreeisnumbered1
第七天堂·2011-10-25 20:00

JOJ1037：Operand

JOJ1037：OperandProfessorMapleteachesmathematicsinauniversity.Hehaveinventedafunctionforthepurposeofobtainingtheoperandsfromanexpression.Thefunctionnamedop
第七天堂·2011-10-25 19:00

joj1435

 1435:GoldbachResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K787206StandardGoldbach'sConjecture: Foranyevennumber n greaterthanorequalto4,thereexistsatleastonepairofprimenumbers p1 and p2 suchth
hechenghai·2011-10-23 23:00

joj1240

 1240:TheDecoderResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K849169StandardWriteacompleteprogramthatwillcorrectlydecodeasetofcharactersintoavalidmessage.Yourprogramshouldreadagivenfileofasimpl
fangzhiyang·2011-10-23 16:00

joj2318

 2318:DiamondResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s16384K1552578StandardDrawadiamondshapewithcharacter'*'.InputThereareseverallinesintheinput,eachlineisaintegerN(N #include usingnamespaces
fangzhiyang·2011-10-23 16:00

joj2201

 2201:PrimeandDecompoundingResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K400209StandardApositiveintergerexcept1canbedecompoundedtothesumofseveralprimenumbers(includeonlyonenumber).Forexsample,f
hechenghai·2011-10-22 23:00

joj1583

 1583:LetMeCountTheWaysResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K551231StandardAftermakingapurchaseatalargedepartmentstore,Mel'schangewas17cents.Hereceived1dime,1nickel,and2pennies.Latertha
hechenghai·2011-10-22 23:00

joj1424

 1424:CoinChangeResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K700243StandardSupposethereare5typesofcoins:50-cent,25-cent,10-cent,5-cent,and1-cent.Wewanttomakechangeswiththesecoinsforagivenamoun
hechenghai·2011-10-22 23:00

joj2474

先找各阶段的状态，再dp规律是：2*1的不同的数目为2；2*2的不同的数目为7；2*3的不同的数目为8；以下全为8，注意这里说的不同的数目是指2*n中横跨n个格的放法，即不可分割的放法。 #include#includeusingnamespacestd;intdp[101];intmain(){   intn,i;   memset(dp,0,sizeof(dp));   dp[1]=2;   
qiankun1993·2011-10-21 21:00

joj2308

 贪心，弄个2重循环，每次取最优就可以了。#include#include#include#includeusingnamespacestd;inta[52];intmain(){   intn,k;   while(scanf("%d%d",&n,&k)!=EOF)   {       inti,j;       for(i=1;i=k)              {              
qiankun1993·2011-10-21 19:00

joj2325

据说要用dp，不过我用的是BFS，就是从上下左右四个方向压缩，在余下的面积不小于max的情况下一直到压缩出全为一的子阵，然后更新max的值，反复进行这个操作，即一开始的矩阵为：1111000011110000可以压缩为：0000 （去掉第一行）11110000   1111 （去掉第四行）00001111  111（去掉第一列）000111000 111（去掉第四列）000111000然后反复压缩
qiankun1993·2011-10-21 17:00

joj 1270

1270:TriangleWaveResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K1339684StandardInthisproblemyouaretogenerateatriangularwaveformaccordingtoaspecifiedpairofAmplitudeandFrequency.InputandOutputThei
fangzhiyang·2011-10-19 18:00

joj1995

 1995:EnergyResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s10240K1336351StandardMr.Jojerisaveryfamouschemist.Heisdoingaresearchaboutbehaviorofagroupofatoms.Atomsmayhavedifferentenergyandenergycanbepo
hechenghai·2011-10-11 23:00

joj2511

 2511:NumbertriangleResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE1s65536K427217Standard5 34 812 5436 21798 AnumbertrangleiscomposedofN(N#includeusingnamespacestd;inta[100+3][100+3];ints[100+3][100+3]
hechenghai·2011-10-07 21:00

JOJ 2474:Tile My Corridor 平铺棋盘 状态压缩解法

dfs找状态没找明白 ，直接手写的状态转移，反正就2^2的情况。#include #include /* constintmod=2008; intf[106]; intmain() { f[0]=1;f[1]=2;f[2]=11;f[3]=7*f[1]+2*f[2]+2; f[4]=f[3]*2+f[2]*7+f[1]*6+f[0]*2; for(inti=5;i=2)
jxy859·2011-10-07 19:00

[置顶] joj神一般的循序渐进200题 扩充版

转载地址：http://blog.csdn.net/jxy859/article/details/6324181Level0(试试吧，电脑判题很神奇的）：AC掉 1000（TheA+BProblem）  吧！(一二阶段，共49题，每天至少三题，由简到难）Level1（格式入门，熟悉OJ）：2484(格式控制)           2357(排序)              1012(数论,最大公约
hechenghai·2011-10-05 20:00

joj2645

http://acm.jlu.edu.cn/joj/showproblem.php?
hechenghai·2011-10-04 20:00

joj2252

[新版judge已支持specialjudge；增加了提交后的状态页面；增加对提交代码的语言检查][新版judge系统近日试运行，希望可以解决内存越界导致Waiting的问题] [more]Welcome, �γɺ�[JPoints: 1207,Sender!: 0/17] 2252:PickBallsResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s
hechenghai·2011-10-04 16:00

joj1751

 1751:UglyNumbersResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K1168351StandardUglynumbersarenumberswhoseonlyprimefactorsare2,3or5.Thesequence1,2,3,4,5,6,8,9,10,12,15,...showsthefirst11uglynumbe
hechenghai·2011-10-04 14:00

joj2653

 2653:不同的数ResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s65536K40982Standard原来有n对数字，每对两个数字都相同。现在其中某对数字中的一个发生了变化。并且所有数字被打乱了顺序。求这一对数字变化后是多少。要求先输出小的。再输出大的。Input多组输入，每组如下：第一行一个数字n,n#includeusingnamespace
hechenghai·2011-10-02 17:00

joj 2645

 并查集的一道题，绞尽脑汁想了用并查集算法但总是过不去，后来在网上查了代码（最近好像总是每次都有一个小错误过不去，蛋疼啊），发现了原来是sum应该定义为longlong--，但我不会用那东西，毕竟不同的oj对他的输入输出控制不同，所以用的double，最后发现竟然挤进了这道前十，对我是莫大的鼓舞啊：#include#include#includeusingnamespacestd;structta
qiankun1993·2011-09-29 12:00

joj2236

 2236:BalanceandPoiseResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE5s8192K311141Standard工匠大师Hill打造出一个精致的天平,对他的徒弟Oren说：“你帮我做一些砝码配套，为了方便携带，砝码数目要尽量少”。Oren：“好。我将按照2进制的规则打造，也就是1，2，4，8，16等等，这样在称量同等重量下使用的砝码数
hechenghai·2011-09-27 18:00

joj2243

 2243:EndlessCarryResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K827306Standard作为2进制的加法，从k加1变成k+1可能会出现若干进位。如1011加1就会有2个进位。给定n，从0开始不停地加1直到n，计算在此过程中总共会有多少次进位。Input输入的每一行有单独的一个值n。n=0标志输入结束Output对于输
hechenghai·2011-09-23 20:00

joj 2536

  计算几何求解最小外接矩形的问题，这道题是暑假训练时做的一道题，当时不会做，选拔赛在即，突然想看看自己进步了多少，便做了一下，想法很容易就有了（说明进步了。。），但总是wa，逼我买了前辈的代码（178point啊），但还是找不到哪错了，最后终于发现最后四个数的初始化有问题，天，我已经初始化为整形最大值了（数据为double类型），还是不够大，天知道他给的测试数据有多变态。。。。话说马上选拔赛了，
qiankun1993·2011-09-23 18:00

joj 1805

求外接圆圆心的一道题#include#include#includeusingnamespacestd;structPoint{      doublex;      doubley;};Point circle_center(Point pt[3]) {           double x1,x2,x3,y1,y2,y3;           double x = 0;           d
qiankun1993·2011-09-22 22:00

joj2218

 2218:StampsResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE5s8192K299151StandardBackground EverybodyhatesRaymond.He'sthelargeststampcollectoronplanetearthandbecauseofthathealwaysmakesfunofalltheothersa
hechenghai·2011-09-22 19:00

joj2170

 2170:TravelResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K495177StandardMikewanttotravelaroundintheholiday,sohefoundsomeinfo.ThetravelstartsatthetimeSandendsatF.Mikewanttodomoretavellingwithout
hechenghai·2011-09-22 19:00

joj1538

 1538:PacketsResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K311108StandardAfactoryproducesproductspackedinsquarepacketsofthesameheight h andofthesizes ,  ,  ,  ,  ,  .Theseproductsarealwaysdeliv
hechenghai·2011-09-22 18:00

joj 1184

   用得非常笨的方法：#include#include#includeusingnamespacestd;#define pi3.141592653589793intmain(){   doublea1,a2,a3,b1,b2,b3,x,y;   while(scanf("%lf%lf%lf%lf%lf%lf",&a1,&b1,&a2,&b2,&a3,&b3)!=EOF)   {       d
qiankun1993·2011-09-21 19:00

joj1048

 1048:WoodenSticksResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K884291StandardThereisapileofnwoodensticks.Thelengthandweightofeachstickareknowninadvance.Thesticksaretobeprocessedbyawoodworkingm
hechenghai·2011-09-20 20:00

joj2487

 2487:CentralAvenueRoadResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s65536K743169StandardTherearealotoftreesintheparkofJilinUniversity.Wewanttoopenuparoadinthisforestinastraightlinefromonetreetoanot
hechenghai·2011-09-18 18:00

joj 1131 Intersection

一道wa了n+1次的题，都快绝望了，本想求助于网络上的各种代码，却发现一律模板，话说我现在还不知道模板为何物。。。。。。。终于在我的绞尽脑汁之下找到了错误，心情very舒畅，方法是看直线和矩形的各个边能不能相交（规范和非规范）以及他会不会在矩形里边：#include#includeusingnamespacestd;structpoint{   intx;   inty;};intdet(inta
qiankun1993·2011-09-18 16:00

joj1611

 1611:PerfectCubesResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K763314StandardForhundredsofyearsFermat'sLastTheorem,whichstatedsimplythatfor n >2thereexistnointegers a, b, c >1suchthat  ,hasrem
hechenghai·2011-09-16 22:00

joj1873

 1873:PowerStringsResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE10s8192K640254StandardGiventwostrings a and b wedefine a*b tobetheirconcatenation.Forexample,if a="abc"and b="def" then a*b="abcdef".Ifw
hechenghai·2011-09-16 22:00

joj1466

 1466:TheHammingDistanceProblemResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K417202StandardTheHammingdistancebetweentwostringsofbits(binaryintegers)isthenumberofcorrespondingbitpositionsthatdif
hechenghai·2011-09-16 20:00

joj 1169

 这道题着实让我蛋疼了一下，看完书上的代码之后信心满满还稍微修改了一下方法来做这道题，结果wa的暗无天日，后来猛然想到oj的测试数据不一定一半是零，然后感觉样例输入果然很坑人，然后AC： #include#include#includeusingnamespacestd;intcap[32][32];intflow[32][32];inta[32];intp[32];intmain(){   in
qiankun1993·2011-09-16 20:00

joj1018

 1018:AnagramsbyStackResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE5s8192K1175519StandardHowcananagramsresultfromsequencesofstackoperations?TherearetwosequencesofstackoperatorswhichcanconvertTROTtoTOR
hechenghai·2011-09-14 20:00

joj1091

 1091:P,MTHBGWBResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K328208StandardMorsecoderepresentscharactersasvariablelengthsequencesofdotsanddashes.Inpractice,charactersinamessagearedelimitedbysho
hechenghai·2011-09-14 20:00

joj1141

 1141:RailsResultTIMELimitMEMORYLimitRunTimesACTimesJUDGE3s8192K867302StandardThereisafamousrailwaystationinPopPushCity.Countrythereisincrediblyhilly.Thestationwasbuiltinlastcentury.Unfortunately,fund
hechenghai·2011-09-14 18:00
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