UVa 457 - Linear Cellular Automata

LinearCellularAutomata

Timelimit:3.000seconds

AbiologistisexperimentingwithDNAmodificationofbacterialcoloniesbeinggrowninalineararrayofculturedishes.BychangingtheDNA,heisable``program"thebacteriatorespondtothepopulationdensityoftheneighboringdishes.Populationismeasuredonafourpointscale(from0to3).TheDNAinformationisrepresentedasanarrayDNA,indexedfrom0to9,ofpopulationdensityvaluesandisinterpretedasfollows:

Inanygivenculturedish,letKbethesumofthatculturedish'sdensityandthedensitiesofthedishimmediatelytotheleftandthedishimmediatelytotheright.Then,bythenextday,thatdishwillhaveapopulationdensityofDNA[K].

Thedishatthefarleftofthelineisconsideredtohavealeftneighborwithpopulationdensity0.

Thedishatthefarrightofthelineisconsideredtohavearightneighborwithpopulationdensity0.

Now,clearly,someDNAprogramscauseallthebacteriatodieoff(e.g.,[0,0,0,0,0,0,0,0,0,0]).Othersresultinimmediatepopulationexplosions(e.g.,[3,3,3,3,3,3,3,3,3,3]).ThebiologistisinterestedinhowsomeofthelessobviousintermediateDNAprogramsmightbehave.

Writeaprogramtosimulatetheculturegrowthinalineof40dishes,assumingthatdish20startswithapopulationdensityof1andallotherdishesstartwithapopulationdensityof0.

Input

Theinputbeginswithasinglepositiveintegeronalinebyitselfindicatingthenumberofthecasesfollowing,eachofthemasdescribedbelow.Thislineisfollowedbyablankline,andthereisalsoablanklinebetweentwoconsecutiveinputs.

ForeachinputsetyourprogramwillreadintheDNAprogram(10integervalues)ononeline.

Output

Foreachtestcase,theoutputmustfollowthedescriptionbelow.Theoutputsoftwoconsecutivecaseswillbeseparatedbyablankline.

Foreachinputsetitshouldprintthedensitiesofthe40dishesforeachofthenext50days.Eachday'sprintoutshouldoccupyonelineof40characters.Eachdishisrepresentedbyasinglecharacteronthatline.Zeropopulationdensitiesaretobeprintedasthecharacter`'.Populationdensity1willbeprintedasthecharacter`.'.Populationdensity2willbeprintedasthecharacter`x'.Populationdensity3willbeprintedasthecharacter`W'.

SampleInput

1

0120133230

SampleOutput

bbbbbbbbbbbbbbbbbbb.bbbbbbbbbbbbbbbbbbbb
bbbbbbbbbbbbbbbbbb...bbbbbbbbbbbbbbbbbbb
bbbbbbbbbbbbbbbbb.xbx.bbbbbbbbbbbbbbbbbb
bbbbbbbbbbbbbbbb.bb.bb.bbbbbbbbbbbbbbbbb
bbbbbbbbbbbbbbb.........bbbbbbbbbbbbbbbb
bbbbbbbbbbbbbb.xbbbbbbbx.bbbbbbbbbbbbbbb
bbbbbbbbbbbbb.bbxbbbbbxbb.bbbbbbbbbbbbbb
bbbbbbbbbbbb...xxxbbbxxx...bbbbbbbbbbbbb
bbbbbbbbbbb.xb.WW.xbx.WW.bx.bbbbbbbbbbbb
bbbbbbbbbb.bbb.xxWb.bWxx.bbb.bbbbbbbbbbb

Note:Wheshowonlythefirsttenlinesofoutput(thetotalnumberoflinesmustbe50)andthespaceshavebeenreplacedwiththecharacter"b"foreaseofreading.TheactualoutputfilewillusetheASCII-spacecharacter,not"b".

#include<stdio.h>
#include<string.h>

intis_char(intw[],intt)
{
intflag=w[t];
switch(flag)
{
case0:printf("");break;
case1:printf(".");break;
case2:printf("x");break;
case3:printf("W");break;
default:return-1;
}
}

intmain()
{

intdish[42],DNA[10],dishcpy[42],i,j,n,midl,t,data;
scanf("%d",&n);
for(t=1;t<=n;++t)
{
memset(dish,0,sizeof(dish));
memset(dishcpy,0,sizeof(dish));
dishcpy[20]=dish[20]=1;dish[0]=0;
memset(DNA,0,sizeof(DNA));
for(i=0;i<10;++i)
scanf("%d",&DNA[i]);
for(data=1;data<=50;++data)
{
for(j=1;j<=40;++j)is_char(dishcpy,j);printf("\n");
for(midl=1;midl<=40;++midl)
{
dishcpy[midl]=DNA[(dish[midl-1]+dish[midl]+dish[midl+1])];
}
for(j=1;j<=40;++j)dish[j]=dishcpy[j];

}
if(t!=n)printf("\n");
}
return0;
}

解题报告：

一开始题目长且对题目所指的意思不太明白，DNA序列是你要输入的数据，所指dish的值=DNA中第K个数的值=所指dish细菌浓度+所指dish左边细菌浓度+所指dish右边细菌浓度（0—40）tip：细菌的浓度（0-3表示），最左边的dish假设还有一个浓度为0的dish，同理右边也是如此假设。输出每天dish值的变化，共50天；input中要求输入要输入空格，可以忽略，正是以为要自己输出blankspace弄得我WA了一次，这点教训要谨记！

一开始看不懂follows中的farlefthasleft，忽略后做出来的题目肯定得WA了，二就是对input表示不理解，换行是否要自己输入还是要自己输出，这点完全不确认，所以注定要WA了，三是没看output后面的Note，这点本来不应该的。总来说，英语的理解能力还待提高，耐心还待培养，思维能力还待提高，最重要的还是书要看、题要做！这题涨了点经验了。