UVa 457 - Linear Cellular Automata
2014-05-03 22:00
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13583110 | 457 | Linear Cellular Automata | Accepted | ANSI C | 0.012 | 2014-05-03 21:48:27 |
翻译见http://www.nocow.cn/index.php/UVA/457
【解决过程】
代码如下:#include<stdio.h>
char con(int c){
if(c==0) return ' '; else
if(c==1) return '.'; else
if(c==2) return 'x'; else
return 'W';
}
int main(){
int i,n;
scanf("%d",&n);
for(i=1;i<=n;i++){
int j,k,DNA[10],dish[50][42]={{0}};
for(j=0;j<=9;j++) scanf("%d",&DNA[j]);
dish[0][20]=1;
for(j=0;j<50;j++){
for(k=1;k<=40;k++)
printf("%c",con(dish[j][k]));
if(j<49) for(k=1;k<=40;k++)
dish[j+1][k]=DNA[dish[j][k-1]+dish[j][k]+dish[j][k+1]];
printf("\n");
}
if(i<n) printf("\n");
}
return 0;
}
【教训】
1、一定要注意数组的范围!
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