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Balloon
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
map
HDU
1004
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
C++
HDU
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
kvm
内存分配
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
ZOJ
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
字符串
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
编程
C语言
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
ACM
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
ACM
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
OO
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
HDU
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
图论
2SAT
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
Crash
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
c
模拟
【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
数据结构
算法
balloon
成鹏致远
Crashing
mystery
【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
算法
数据结构
成鹏致远
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
com
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
java
开发技术
【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
String
iterator
存储
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
String
iterator
input
UP
each
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
jquery
function
datagrid
table
Class
div
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
String
iterator
include
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
c
python
haskell
ACM
DFS
hdoj 1004 Let the
Balloon
Rise(stl)
LettheBalloonRiseTimeLimit:2000/1000MS(Java/Others) MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):41590 AcceptedSubmission(s):14543ProblemDescriptionContesttimeagain!Howexciteditistose
hnuzengchao
·
2012-06-20 10:00
String
iterator
input
UP
each
IOS实现图片倒影效果
[superviewDidLoad]; [[[selfview]layer]setBackgroundColor:[[UIColorblackColor]CGColor]]; UIImage*
balloon
zhibudefeng
·
2012-06-19 17:00
ios
import
layer
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