csu 1395: Timebomb (模拟)
题意：给出若干数字 如果能被6整除 输出 BEER!! 否则输出 BOOM!!
思路：用三维数组存下0到9
暴力出每个数的值
求出结果
（这题有点坑在于 1 数字可能不存在 2 数字的数量不确定）
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char num[10][10][10]=
{
{"***","* *","* *","* *","***"},
{" *"," *"," *"," *"," *"},
{"***"," *","***","* ","***"},
{"***"," *","***"," *","***"},
{"* *","* *","***"," *"," *"},
{"***","* ","***"," *","***"},
{"***","* ","***","* *","***"},
{"***"," *"," *"," *"," *"},
{"***","* *","***","* *","***"},
{"***","* *","***"," *","***"}
};
char mat[100][100];
int main()
{
int i,j,k;
int l,m,n;
int sum,res;
memset(mat,0,sizeof(mat));
while(gets(mat[0]))
{
sum=0;res=1;
for(i=1;i<=4;i++)
{
gets(mat[i]);
}
int len=strlen(mat[0]);
for(i=0;i<len;i+=4)
{
for(j=0;j<10;j++)
{
/*if(j==6)
{
for(k=0;k<5;k++)
{
for(l=0;l<3;l++)
{
printf("%c",num[j][k][l]);
}
printf("\n");
}
}*/
for(k=0;k<5;k++)
{
for(l=0;l<3;l++)
{
if(num[j][k][l]=='*')
{
if(mat[k][l+i]!=num[j][k][l]) { break;}
}
else if(num[j][k][l]==' ')
{
if(mat[k][l+i]=='*') break;
}
}
if(l!=3) { break;}
}
//if(j==6) printf("%d....\n",k);
if(k==5)
{
break;
}
}
//printf("%d...\n",j);
if(j<10)
{
sum=sum*10+j;
}
else
{
res=0;
break;
}
}
//printf("%d %d\n",res,sum);
if(res==0||sum%6!=0)
printf("BOOM!!\n");
else
printf("BEER!!\n");
memset(mat,0,sizeof(mat));
}
return 0;
}