balloon

HDU1170 Balloon Comes!

BalloonComes!TimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):19874    AcceptedSubmission(s):7486ProblemDescriptionTheconteststartsnow!Howexciteditistoseeb
u012846486·2014-08-28 08:00

HDU1004 Let the Balloon Rise tire树

题目大意：找出若干字符串中出现频率最高的字符串并输出。分析：Tire树的应用，不过本题数据量比较小，纯暴力也能A。Tire树实现代码如下：#include #include #include usingnamespacestd; #defineson_num30//字符串中包含的字符个数 #definemaxn15//单词的最大长度 structtire { intnum;//纪录到达该节点的字符
AC_Gibson·2014-08-07 10:00

HDU 1004 Let the Balloon Rise

pid=1004   Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    
hellojyj·2014-07-31 10:00

1407261735-hd-Let the Balloon Rise.cpp

#include#includeintmain(){ chars[1000][100],a[1000]; intn; inti,j,k,max; while(scanf("%d",&n),n) {  getchar();  for(i=0;i
wangluoershixiong·2014-07-26 18:00

1004Let the Balloon Rise

#include #include #include #include usingnamespacestd; structnode { stringst;intcnt; node(strings,intk=1):st(s),cnt(k){} node(){} booloperator>n,n) { strings; vectorball; vector::iteratorit; for(inti=
u013827143·2014-07-21 12:00

Hdu 1004 Let the Balloon Rise map解决方法

原题地址http://acm.hdu.edu.cn/showproblem.php?pid=1004#include #include #include #include usingnamespacestd; mapm; intmain() { chars[1005][20]; intn; while(~scanf("%d",&n)&&n!=0) { getchar(); m.clear(); i
tenlee·2014-07-15 19:00

HDU 1004 Let the Balloon Rise

#include #include #include usingnamespacestd; intmain() { charstr[1050][20]; inta[1050],i,j,k,n; while(cin>>n,n) { getchar(); memset(a,0,sizeof(a)); intmax=0; for(i=0;i0) { for(j=0;jmax) max=a[j]; } }
u014142379·2014-05-09 12:00

kvm的气泡（balloon）机制 及 在线资源调整

最近想借用kvm虚拟化中的balloon机制来实现linux云主机的在线资源伸缩，由于cloudstack也采用balloon技术来在线调整云主机的资源，因此在可行性上应该没有问题，不过在实际测试中碰到了奇怪的现象
weiyuanke·2014-04-15 21:00

hdu 1170 Balloon Comes!（水题）

小记：PE了我几次，看不懂。可能是读入字符串读了空串，然后就会多输出一个换行，这样才导致的PE。思路：switch。注意除法，如果能整除就不留小数。代码：#include #include #include usingnamespacestd; #defineN100010 #defineMAX1000010 intT,a,b,n; charc[10]; intmain(){ while
ljd4305·2014-04-09 16:00

ZOJ 1003 Crashing Balloon

题目来源：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3这是一道模拟题！#include #include #include usingnamespacestd; boolaTrue,bTrue; /** 判断a,b有没有公共的因子，如果有，那么分数低的胜，如果没有，那么分数高的获胜 如果分数在2-100之间，那么就不用
Hearthougan·2014-04-02 15:00

模拟 --- 字符串统计

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536
·2014-03-24 12:00

hdu Let the Balloon Rise

LettheBalloonRiseTimeLimit:2000/1000ms(Java/Other)   MemoryLimit:65536/32768K(Java/Other)TotalSubmission(s):63   AcceptedSubmission(s):26Font:TimesNewRoman|Verdana|GeorgiaFontSize:←→ProblemDescription
u014028231·2014-02-16 14:00

HDOJ Let the Balloon Rise

DescriptionContesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoonsofeachcolor
u013013910·2014-01-28 20:00

Let the Balloon Rise

ProblemDescriptionContesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoonsofea
u010257696·2013-11-27 08:00

即时通知

现在已经有很多工具能够帮助实现这个过程，包括NuGet和VisualStudio2012中所使用的气球（balloon）通知。每一种实现方式都有它自己的缺点。
Jeff Martin·2013-09-24 00:00

[catch]--Let the Balloon Rise

ProblemDescriptionContesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoonsofea
a191030148·2013-09-13 08:00

Let the Balloon Rise

#include usingnamespacestd; intmain() { intnum; while(cin>>num) { stringtemp; inti,j; intcount=0; intmax=0; if(num==0)break; stringcolor[num]; for(i=0;i>color[i]; for(i=0;imax) { m
u010257696·2013-09-06 09:00

[ACM]Let the Balloon Rise

ProblemDescriptionContesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoonsofea
sr19930829·2013-09-01 12:00

hdu1004Let the Balloon Rise

#include #include #include #include #include usingnamespacestd; //2013-08-3115:48:21Accepted10040MS372K956BG++Achiberx intmain() { intn; stringt; mapmymap; map::iteratorit; while(scanf("%d",&n)!=EOF&
wangwenhao00·2013-08-31 15:00

HD1004Let the Balloon Rise

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others
·2013-08-20 15:00

zoj 3717 - Balloon(2-SAT)

裸的2-SAT，详见刘汝佳训练指南P-323不过此题有个特别需要注意的地方：Youshouldpromisethatthereisstillnooverlapforanytwoballoonsafterrounded.模版题，代码如下：#include #include #include #include #include #include #include #include #include #
shankeliupo·2013-08-20 08:00

HDU1004——Let the Balloon Rise

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest
·2013-08-14 19:00

ACM-水题之 let the balloon rise——hdu1004

LettheBalloonRiseTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):65180    AcceptedSubmission(s):24152ProblemDescriptionContesttimeagain!Howexciteditistose
lx417147512·2013-07-30 09:00

zoj3717 Balloon(二分+2SAT)

BalloonTimeLimit: 3Seconds     MemoryLimit: 65536KB     SpecialJudgeTheweatheriswonderfultoday. Gao takesawalkinthegardenwithhisgirlfriend.Hisgirlfriendlikesballoonssomuch,sothatshewantstoflysomeballo
ophunter·2013-07-23 10:00

HDU 1004 Let the Balloon Rise

题目链接:LettheBalloonRise解题思路:直接用STL就可以做出来，用map的映射，形成一对一的关系。#include #include #include #include usingnamespacestd; intmain(){ stringasd; charcolor[20]; inti,n,max,index; while(scanf("%d",&n)&&n){ max
u010787640·2013-07-21 21:00

hdu-Balloon Comes!

hdu-BalloonComes!一道水题的报告今天闲来无事，想刷几道水题来增加自信心。但是在刷这道水题时，我发现了自己的一个知识漏洞！！！！ProblemDescriptionTheconteststartsnow!Howexciteditistoseeballoonsfloatingaround.You,oneofthebestprogrammersinHDU,cangetaverybeaut
yinzm520·2013-07-19 22:00

HDU 1004 Let the Balloon Rise

LettheBalloonRiseTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):56685    AcceptedSubmission(s):20670ProblemDescriptionContesttimeagain!Howexciteditistose
fjy4328286·2013-07-12 08:00

ZOJ 3717 Balloon 解题报告

 BUPTSummertraining1题意：现有n组气球（气球可以看成球体），每组有一个红气球一个蓝气球，已知每组气球的球心。要求从这n组气球中每组挑出一个气球放入花园中，不允许任意两球之间有重叠，求满足条件的气球的最大半径。解法：因为每组就两个球，很容易想到二分+2sat。      首先二分枚举半径R。然后建图，若球x与球y的球心距小于R，则x->y'，y->x'连边。跑一遍2sat，并判断
u010638776·2013-07-04 14:00

ZOJ 3717: Balloon

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3717题意：有N组气球，每组有一个蓝气球以及一个红气球可供选择，它们各自有自己的圆心坐标。在每组中选择一个气球吹起来，要求把所有的气球吹到一样大，气球和气球之间不能重叠。问最多可以吹多大。算法：典型的二分求值+2-SAT判定。o(n^2)预处理出距离。每次判定，若两
frog1902·2013-06-30 15:00

ZOJ 3717 Balloon (2sat+二分)

这题是一个二分半径+判定的题，但是开始不知道怎么去判定，以前听别人提起过sat问题，感觉这题是一个2sat模型，因为有一些限制比如：同一组两个里面只能选一个，两个球重叠时只能选其中的一个等，这很符合2sat的模型，于是去翻了下白书上2sat问题的解法实现，建图后直接用搜索判定，我用这种方法写了写，开始提交wa了几次，wa的原因是：题目说在round之后也要保证不重叠，那就只有舍掉小数点3位以后的数
laziercs·2013-06-30 13:00

【Acm】算法之美—Crashing Balloon

题目概述：CrashingBalloon Onevery June1st,theChildren'sDay,therewillbeagamenamed"crashingballoon"on TV. Theruleisverysimple. Onthegroundthereare100labeledballoons, withthenumbers1to100.Aftertherefereeshout
·2013-06-27 17:00

HDU OJ 1004 Let the Balloon Rise

HDUOJ1004LettheBalloonRiseLettheBalloonRiseTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):60483AcceptedSubmission(s):22247ProblemDescriptionContesttimeagain!
RePorridge·2013-06-12 19:00

HDU 1004 Let the Balloon Rise

                                   LettheBalloonRiseTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):55611    AcceptedSubmission(s):20193ProblemDescription
lsh670660992·2013-05-30 14:00

【算法】算法之美―Crashing Balloon

题目概述：CrashingBalloonOnevery June1st,theChildren'sDay,therewillbeagamenamed"crashingballoon"on TV. Theruleisverysimple. Onthegroundthereare100labeledballoons, withthenumbers1to100.Aftertherefereeshouts
infohacker·2013-05-07 22:25

【算法】算法之美—Crashing Balloon

题目概述：CrashingBalloonOneveryJune1st,theChildren'sDay,therewillbeagamenamed"crashingballoon"onTV.Theruleisverysimple.Onthegroundthereare100labeledballoons,withthenumbers1to100.Aftertherefereeshouts"Let'
infohacker·2013-05-07 22:25

hdu 1170(Balloon Comes!)

#include<stdio.h> #include<math.h> #define zero 1e-10 int main() { int _case; char ys; int a,b,c; double d; while(scanf("%d",&_case)) {
·2013-04-25 16:00

2104_Let the Balloon Rise

Contesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoonsofeachcolorandfindther
asongsongsong·2013-04-20 18:00

ZOJ 1003 Crashing Balloon 搜索

题意：100个气球，气球上标有1-100的号码，每踩一个气球，则自己的得分可以乘以该气球的标号（初始得分为1，每个气球只能踩一次）。题解：假如a>b，求出a，b所有可能的分解情况（分解为1-100的数的乘积）。然后比对，只要存在一种分解情况，使得a的因子中不含b的因子，a就是可能的。#include #include #include #include usingnamespacestd; #d
Tsaid·2013-04-15 21:00

hdu1004――Let the Balloon Rise

 原题：ProblemDescriptionContesttimeagain!Howexciteditistoseeballoonsfloatingaround.Buttotellyouasecret,thejudges'favoritetimeisguessingthemostpopularproblem.Whenthecontestisover,theywillcounttheballoons
bingsanchun·2013-03-31 22:00

【map热手题】HDU 1004—Let the Balloon Rise

题目：点击打开链接建议初学STL的人用MAP来切这道水题，可以对MAP的方便之处有更加深刻的理解。因为调用和查找的确挺简单的。效率据查阅资料，是lgN.#include #include #include usingnamespacestd; intmain() { intballnum; while(cin>>ballnum&&ballnum!=0) { stringtemp; intbigge
mig_davidli·2013-01-21 14:00

Hdu 1004 - Let the Balloon Rise

字符串匹配本题要求出ACM比赛现场的最热门的题目所对应的气球颜色，用字符串匹配的思想做。  AC代码：#include #include #include intmain() { chars[1001][16],k[16]; intn,i,j,sum,max; while(scanf("%d",&n),n) { for(i=0;i
Chuck_0430·2012-11-06 11:00

HDOJ 1170 Balloon Comes!

#include int main() { intt; inta,b; chars; scanf("%d",&t); while(t--){ getchar(); scanf("%c%d%d",&s,&a,&b); if(s=='+')printf("%d",a+b); elseif(s=='-')printf("%d",a-b); elseif(s=='*')printf("%d",a*b);
電泡泡·2012-11-04 21:00

1004let the balloon rise

1:用容器和迭代器做：注意：代码提交时可以有注释，但是格式一定要正确，另外不能有多余的输出，只能有题目要求的结果#include#include#includeusingnamespacestd;intmain(){  intn;  while(cin>>n,n!=0)  {    vectorsvec;    vector::iteratoriter1=svec.begin();    vect
xuexiacm·2012-10-27 09:00

HDOJ 1004 Let the Balloon Rise (map)

http://acm.hdu.edu.cn/showproblem.php?pid=1004题意：求出现次数最多的颜色。思路：用map存储颜色和出现次数，然后遍历一遍查找出现次数最多的即可。#include #include #include usingnamespacestd; intmain() { intn; mapcolor; while(cin>>n&&n) { color.clea
sdc1992·2012-09-07 23:00

hdu 1004 Let the Balloon Rise map的应用

LettheBalloonRiseTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):44434    AcceptedSubmission(s):15730ProblemDescriptionContesttimeagain!Howexciteditistose
hnust_xiehonghao·2012-08-26 23:00

HDOJ 1004 Let the Balloon Rise

開源中國寫的第一篇日誌加油水呀水，不過水壓不小，暈死，一晚上就死在了這裡，回想一下這題真的真的不難一開始是思路問題，到後來編程了代碼實現的問題一開始是想的列出兩個數組，一個color[10001][16]存放balloon
電泡泡·2012-08-24 23:00

使用easyUI创建一个拖放的购物车

查看Demo显示商品在页面: Balloon Price:$25 Feeling Price:$25
yhc13429826359·2012-08-08 15:00

hdu 1004.Let the Balloon Rise

#include#include#include#includeusingnamespacestd;mapcolor;intmain(){  intn,max;  stringstr;  map::iteratoriter;  while(cin>>n&&n)  {    color.clear();    max=0;    while(n--)    {      cin>>str;     
lezong2011·2012-07-26 15:00

ios 图片倒影投影的原理详解

就写一下自己在投影上面学到的东西吧原始图片： [[[selfview]layer]setBackgroundColor:[[UIColorblackColor]CGColor]]; UIImage*balloon
x1135768777·2012-07-07 21:00

谁主张、谁举证 ACM crashing balloon / zoj 1003

题目详情可以参考这里： http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3 ZOJ上的判定标准是： b不服，站出来质疑； 如果a能举证说：你瞧，存在一种合理的解释，a = a[1]*a[2]*…*a[n],  b = b[1]*b[2]*…*b[m]; 其中 2<= a[i], b[j] <=1
liuxinyu95·2012-07-03 18:00
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