457 - Linear Cellular Automata

LinearCellularAutomata
AbiologistisexperimentingwithDNAmodificationofbacterialcoloniesbeinggrowninalineararrayofculturedishes.BychangingtheDNA,heisable``program"thebacteriatorespondtothepopulationdensityoftheneighboringdishes.Populationismeasuredon
afourpointscale(from0to3).TheDNAinformationisrepresentedasanarray
DNA,indexedfrom0to9,ofpopulationdensityvaluesandisinterpretedasfollows:

Inanygivenculturedish,letKbethesumofthatculturedish'sdensityandthedensitiesofthedishimmediatelytotheleftandthedishimmediatelytotheright.Then,bythenextday,thatdishwillhaveapopulationdensityof
DNA[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]).ThebiologistisinterestedinhowsomeofthelessobviousintermediateDNAprograms
mightbehave.

Writeaprogramtosimulatetheculturegrowthinalineof40dishes,assumingthatdish20startswithapopulationdensityof1andallotherdishesstartwithapopulationdensityof0.

Input

Theinputbeginswithasinglepositiveintegeronalinebyitselfindicatingthenumberofthecasesfollowing,eachofthemasdescribedbelow.Thislineisfollowedbyablankline,andthereisalsoablanklinebetweentwoconsecutiveinputs.

ForeachinputsetyourprogramwillreadintheDNAprogram(10integervalues)ononeline.

Output

Foreachtestcase,theoutputmustfollowthedescriptionbelow.Theoutputsoftwoconsecutivecaseswillbeseparatedbyablankline.

Foreachinputsetitshouldprintthedensitiesofthe40dishesforeachofthenext50days.Eachday'sprintoutshouldoccupyonelineof40characters.Eachdishisrepresentedbyasinglecharacteronthatline.Zeropopulationdensitiesaretobeprinted
asthecharacter`'.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>
#include<stdlib.h>
intdish[42],temp[42];

voidprint(){
inti;
for(i=1;i<=40;++i)
{
switch(dish[i])
{
case0:printf("");break;
case1:printf(".");break;
case2:printf("x");break;
case3:printf("W");break;
}
}
printf("\n");
}

intmain(){
intDNA[10],SIZE=sizeof(dish);
intN,i,j;

scanf("%d",&N);

while(N--){
for(i=0;i<10;++i)
scanf("%d",&DNA[i]);

for(i=0;i<=41;++i)
dish[i]=0;
dish[20]=1;

for(i=1;i<=50;++i)
{
print();
for(j=1;j<=40;++j)
temp[j]=DNA[dish[j]+dish[j-1]+dish[j+1]];
memcpy(dish,temp,SIZE);
}

if(N)//必不可少
printf("\n");
}

return0;

}