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;
}