CSU 1395:timebomb (Analog)

Test instructions: Give some numbers if they can be divisible by 6 output beer!! Otherwise output boom!!

Idea: Save 0 to 9 with a three-dimensional array

Violence out of every number of values

Find the results

(This is a bit of a hole in the 1 number may not exist 2 number of uncertainties)

#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) {to (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;}

　　

CSU 1395:timebomb (Analog)