Machined

Getting Started 414 - Machined Surfaces

MachinedSurfacesAnimagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthe
鸿麒·2020-08-23 05:52

UVa - 414 - Machined Surfaces(AC)

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
sophomoreCDD·2020-08-13 18:33

UVA - 414 Machined Surfaces

MachinedSurfacesTimeLimit:3000MSMemoryLimit:Unknown64bitIOFormat:%lld&%lluSubmitStatusDescriptionAnimagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheac
ACM弱渣的平凡之路·2020-08-13 12:07

UVA - 414 Machined Surfaces（题意不好懂）

懂了题意就是水题。给你n行数每行都有25个字符，第一个和最后一个都是‘X'，中间可能包含空格，问每次与左边（或者右边）移动，直到中间没有空格为止，问此时还有多少空格。我们直接考虑移动完之后的状态，肯定是空格少的那一行最先移动完，那么只要统计出原先总的空格数减去空格最少的那一行乘×n就是目前剩下的字符了。#include#includeintmain(){intn,i,j,m,sum,maxn;ch
NowAndForever·2020-08-13 10:35

414 - Machined Surfaces

MachinedSurfacesAnimagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthe
weixin_30784501·2020-08-10 23:12

uva414 - Machined Surfaces

uva414-MachinedSurfaces/*水题，值得一提的是，getline使用时注意不能让它多吃回车键，处理方法可以用getchar。*/#include#include#includeusingnamespacestd;intmain(){intn;while(cin>>n,n){getchar();//第一个回车键会被getline拿去，所以要用getchar处理这个回车定义在cst
weixin_30421525·2020-08-10 22:08

POJ 1493 Machined Surfaces（我的水题之路——移动后的空格）

MachinedSurfacesTimeLimit:1000MSMemoryLimit:10000KTotalSubmissions:1712Accepted:1123DescriptionAnimagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheacho
只为你晴天·2020-08-10 21:33

UVA-414-Machined Surfaces

这道题目就是以水题，无非就是题意难以理解而已。题目的大致意思是：首先输入一个数字N，代表接下来要输入几行字符串；每行字符串的格式为：由字母X和空格（“”）组成，且两边是字符X，中间是空格。每行的开始有一个或多个X，中间有0个或多个空格，末端有一个或多个X。要求：将N行中的左边的字符X同时向右移动（或右边的字符X向左移动），直到有一行中间没有空格时停止所有行的移动，此时计算N行中总共有多少个空格。大
青竹梦·2020-08-10 20:34

UVA414 - Machined Surfaces

问题描述输入多行，每行25个字符，左右为X，中间为空格，当将左边的X平移至与右侧X相连后停止，求平移后之间存在几个空格。思路ans=总空格数-行数*每行中空格最少的数。注意数据中的B是文中的空格，需要替换。代码#include#include#includeusingnamespacestd;intmain(){intminn,n,i,j,s,space;chara[15][30];while(c
听峰问雨·2020-08-10 18:47

UVa 414 - Machined Surfaces

题目：n个由X和空格组成的串，两边有至少一个X，将n个串压缩，每次每行消除一个空格，问到不能消除时剩余的空格。分析：简单题。统计全体空格数sum_b和最少空格数min_b，则结果就是sum_b-n*min_b。注意：利用gets或者getline读入串。#include#include#include#includeusingnamespacestd;chardata[30];intmain(){
小白菜又菜·2020-08-10 18:35

UVa - 414 - Machined Surfaces 题解

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
iteye_18800·2020-08-10 17:55

UVaOJ 414 - Machined Surfaces

#include#includeintmain(){charbuf[30];memset(buf,'\0',sizeof(buf));intn=0;while(~scanf("%d",&n)&&n){fgets(buf,sizeof(buf),stdin);intsum=0;intmin=65535;for(inti=0;i
IceHe何志远·2020-08-10 17:24

uva 414 - Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
滑头鬼之亨·2020-08-10 13:04

UVA 414(Machined Surfaces)字符串水题

给女朋友写题解QAQ题目链接：https://vjudge.net/problem/UVA-414题意：输入n个由X和空格组成的字符串（X在两端，空格在中间），每次可以同时删除n个字符串中间的空格，当其中一个字符串无法删除（即删到只剩下X）时停止，此时输出剩下的空格数。求出n个字符串中空格最少的字符串，当这个字符串删完全部空格之后，必将停止删除操作，此时总共删除的空格数为（总空格数-n*最少空格数
Eknight123·2020-08-10 13:49

UVA 414 - Machined Surfaces

题意：第一行输入一个n，后面输入n行字符串，每行在左右x之间都存在一些空格，每次操作都使每行空格数减1，直到其中一行没有空格为止，求剩余的总空格数解法：可简化为求总空格数LongSum和最少空格数MinLong，结果即为LongSum-n*MinLong代码如下：#includeusingnamespacestd;intmain(){intn,Long,MinLong,LongSum;charx[
liuxinyu666·2020-08-10 12:52

巧解表达式展开问题

写出来大概是这个样子GroupA:MachineA,MachineB,MachineCGroupB:MachineD,MachineEGroupC:MachineA,MachineCGroupD:GroupB
战神猴哥·2019-11-08 01:46

mongo ObjectID 组成

‘_id’是mongodbObjectID类型的，它由12位结构组成，包括timestamp,machined,processid,counter等。
black_cheng·2018-11-19 19:14

docker-mac安装docker

brewinstalldockerbrewinstalldocker-machine二、创建docker-machined
freshghost1234·2018-05-17 15:19

UVA Machined Surfaces

题意：这道题我读了很久，也没有读懂最后看的解体报告才懂得题意，题目不难，但是还是错了两次，几个字符窜，左边的‘x’向右边移动当和右边的‘x’连接时候，求剩下的字符窜还有几个空格 分析：这题错的原因是当吧B换成空格后scanf没有换成gets想当然的直接提交了，可定wa #include<stdio.h> #include<string.h> const int
·2015-11-13 15:46

414 - Machined Surfaces

#include<stdio.h> #include<string.h> char surface[26]; int main() {  int i,j,k,l,n,total,min;  while(scanf("%d",&n)!=EOF&&n!=0)  {   
·2015-11-11 06:43

UVA 414 - Machined Surfaces

Machined Surfaces  An imaging device furnishes digital images of two machined surfaces that eventually
·2015-11-02 14:33

uva414 - Machined Surfaces

uva414 - Machined Surfaces   /* 水题，值得一提的是，getline使用时注意不能让它多吃回车键，处理方法可以用getchar。
·2015-10-31 09:04

poj1493Machined Surfaces

Machined Surfaces Time Limit: 1000MS   Memory Limit: 10000K Total Submissions
·2015-10-21 11:04

POJ 1493 Machined Surfaces（水~）

Description每张照片由n行串组成，每行串长度为25，都由X开头，空格中间(也可没有)，X结尾。所有串的左边X部分同时右移，直到有一个串没有空格。问这时所有串合起来总共有几个空格Input多组用例，每组用例第一行为字符串数量n，其后是n个字符串Output对于每组用例，输出最后空格数SampleInputSampleOutput400Solution水题其实本质就是统计一下每行空格数字，找
V5ZSQ·2015-08-25 10:00

UVA 414-Machined Surfaces

#include intmain() { charc; intt; while(~scanf("%d",&t)) { if(t==0) break; getchar(); inti,j; intsum=0,min1=10000,min2; for(i=0;imin2?min2:min1; } printf("%d\n",sum-min1*t); } return0; }
Griffin_0·2015-08-19 15:00

UVA题目分类

School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined
·2014-10-24 19:00

UVA - 414 Machined Surfaces

 MachinedSurfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedoft
HelloWorld10086·2014-07-30 15:00

UVA 414 Machined Surfaces

#include intmain(){ intn; while(scanf("%d",&n),n){ getchar(); intmin=100,sum=0,num=n; while(n--){ intcount=0; charstr[30]; gets(str); for(inti=0;i<25;i++) if(str[i]=='') count++; if(count
kl28978113·2014-07-28 19:00

2013秋13级预备队集训练习1 F - Machined Surfaces

 MachinedSurfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedoft
u013015642·2013-12-09 08:00

Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters, "
u013013910·2013-12-04 21:00

poj 1493 Machined Surfaces

这道题的难点在于读题，题读懂了，基本就是很水很水的题目。。。题意：每张照片由n行串组成，每行串长度为25，都由X开头，空格中间(也可没有)，X结尾。所有串的左边X部分同时右移，直到有一个串没有空格。问这时所有串合起来总共有几个空格。 思路：求最短的空格数min，每一行的空格数减去min之和就是结果了。AC的代码：#include intblankNum[15]; charimg[30]; in
Scythe666·2013-11-04 09:00

UVa 414 Machined Surfaces (water ver.)

414-MachinedSurfacesTimelimit:3.000secondshttp://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=94&page=show_problem&problem=355Animagingdevicefurnishesdigitalimagesoftwomachin
synapse7·2013-09-13 16:00

UVA414―― Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
bingsanchun·2013-08-07 20:00

UVA414—— Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
bingsanchun·2013-08-07 20:00

8.414 - Machined Surfaces

XXXXBBBBBBBBBBBBBBBBXXXXX16XXXBBBBBBBBBBBBBBBXXXXXXX15XXXXXBBBBBBBBBBBBBBBBXXXX16XXBBBBBBBBBBBBBBBBBXXXXXX17XXXXBXXXXX1XXXXXXXXXX2XXXXXBXXXX1XXBBXXXXXX0intmain(){  intn,len[15];  while(scanf("%d",&n)!
PandaCub·2013-06-16 00:00

414 - Machined Surfaces

注:以下提到的B都是代指空格.题意:题目本身不难,但是本人英语实在太差,读了半天愣是没能理解题意,在网上看了解释后,题目就很快解出来了.把输入的N行XB组合中的连续X进行平移, 从而压缩B的存在(从左向右平移,每一行与下一行是当成连通的一行进行处理的),当其中一个连续B被压缩完毕时,问这N行XB组合中还剩下多少个B?思路:取出每行B中最小的个数,把其他行的B的个数减去这个最小的B得到的差的和即为所
sailtseng·2013-06-10 14:00

UVA 414 - Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated. Adigitalimageiscomposedofthetwocharacters,"
hellobin·2013-02-26 16:00

寒假刷题之4——机械化表面？

 Machined Surfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedof
hcbbt·2013-01-30 23:00

UVaOJ 414 - Machined Surfaces

AOAPCI: BeginningAlgorithmContests(RujiaLiu) ::Volume0.GettingStartedDescription给你一堆长度都为25的字符串。像样例中，这堆字符串的两边都有X。将两边的X拼起来，就像牙齿咬合一样。因为两边的X数量可能不同，最后拼起来可能会有一些空隙。计算这些空隙的数量。TypeWaterAnalysis统计一共有多少空格，和每行最少有
Ra_WinDing·2012-09-07 22:00

uva 414 Machined Surfaces

小结： 1 英语太烂，看不懂意思，百度的，汗 2 语法不好，不能定义如count[a],之类的，以后用const吧 3用for循环时，i #include intmain(){ inta,i,j; while(scanf("%d",&a)==1){ if(!a)break; chars[15][30]; intcount[15],sum=0; memset(count,
oceaniwater·2012-09-02 21:00

uva 414 - Machined Surfaces

Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedofthetwocharacters,"X
Frankiller·2012-06-01 16:00

UVA-414-Machined Surfaces

这道题目就是以水题，无非就是题意难以理解而已。题目的大致意思是：首先输入一个数字N，代表接下来要输入几行字符串；  每行字符串的格式为：由字母X和空格（“”）组成，且两边是字符X，中间是空格。每行的开始有一个或多个X，中间有0个或多个空格，末端有一个或多个X。  要求：将N行中的左边的字符X同时向右移动（或右边的字符X向左移动），直到有一行中间没有空格时停止所有行的移动，此时计算N行中总共有多少个
rowanhaoa·2012-05-19 14:00

414 - Machined Surfaces

 MachinedSurfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedoft
zhuyi2654715·2012-05-14 20:00

Uva 414 Machined Surfaces

#include #include intmain() { intn,i,j,min,sum; intcount[20]; while(scanf("%d",&n),n) { charch; sum=0; memset(count,0,sizeof(count)); for(i=0;i
ssun125·2012-03-29 15:00

POJ1493 Machined Surfaces

题意：就是将中间的空格合并，问最后剩下几个空格。直接看范例就懂了，看题目描述还比较蛋疼。思路：其实本质就是统计一下每行空格数字，找出最小值。然后剩的值即是sum[每行空格数-最小值]。#include #include #include #include #definemax(a,b)(a>b?a:b) #defineabs(a)((a)>0?(a):-(a)) #definemin(a,b)(a
wuyanyi·2012-02-02 13:00

uva414 Machined Surfaces

 MachinedSurfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedoft
x954818696·2011-11-24 22:00

uva414 Machined Surfaces

 MachinedSurfaces Animagingdevicefurnishesdigitalimagesoftwomachinedsurfacesthateventuallywillbeassembledincontactwitheachother.Theroughnessofthisfinalcontactistobeestimated.Adigitalimageiscomposedoft
·2011-11-24 22:00

UVa 414 - Machined Surfaces

题目大意：有n行字符串，每行中间是空格，两边是字母X，把右边的整体往左边平移，直到某一行中间没有空格就计算此时整个平面里的空格数。首先想到的思路就是：利用两个数组left，right分别记录每一行左边x的终结位置和右边X的开始位置，然后利用right[i]的递减来模拟左移，最终由right[i],left[i]的差值来判断某一行是否已满，最后再把差值的和打印出来就得到结果。有个特殊情况，就是中间没
gneveek·2011-09-27 08:00

UVa 414 Machined Surfaces

UVa414MachinedSurfaces题目大意：不断地消除空格，使左右两边接触，求最后剩下多少个空格。以下是我的代码：#includebool ok(long n,long a[],long b[]){    for(long i=0;i=b[i]-1)        return true;    return false;}int main(){    const long maxn=2
心如止水·2010-01-06 20:00
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