UVA 232-crossword Answers

The code is very confusing. There is no time to tidy up the time too tight, not a good idea ~ but it can be AC ...

I can not see the news, or add QQ joint discussion

#include <stdio.h> #include <string.h> #include <iostream>using namespace Std;char s[11][11];int num[    11][11];int Vis[11][11];int Main () {int r,c;    int t=0;        while (scanf ("%d", &r)!=eof) {if (r==0) break;        t++;        scanf ("%d", &c);        for (int i=0;i<r;i++) {scanf ("%s", S[i]);        }//number; memset (num,0,sizeof (num));        memset (vis,0,sizeof (VIS));        int t1=0;                for (int i=0;i<r;i++) {for (int j=0;j<c;j++) {if (s[i][j]!= ' * ')                    {if (i==0) {num[i][j]=++t1;                    } else if (j==0) {num[i][j]=++t1; } else if (s[i][j-1]== ' * ' | |                    s[i-1][j]== ' * ') {num[i][j]=++t1; }                }            }        }for (int i=0;i<r;i++)//{//for (int j=0;j<c;j++)//{//printf ("        %d ", num[i][j]);//}//printf (" \ n ");//}//Output.        if (t!=1) printf ("\ n");        printf ("Puzzle #%d:\n", t);        printf ("across\n");        int flag=0;        int cnt1=0;        int flag1=1; for (int i=0;i<r;i++) {for (int j=0;j<c;j++) {if (s[i][j]== ' * ' &&f                    LAG) {flag=0;                printf ("\ n");                        } if (s[i][j]!= ' * ') {if (num[i][j]!=0) {                        if (i==0&&num[i][j-1]==0) printf ("%3d.", Num[i][j]); else if (i!=0&& (j==0| |                    s[i][j-1]== ' * ')) printf ("%3d.", Num[i][j]); } if (j==c-1) {                        printf ("%c\n", S[i][j]);                    flag=0;                        } else {printf ("%c", S[i][j]);                    flag=1;        }}}} printf ("down\n");                for (int i=0;i<r;i++) {for (int j=0;j<c;j++) {if (num[i][j]!=0)                    {if (i==0) printf ("%3d.", Num[i][j]);                else if (s[i-1][j]== ' * ') printf ("%3d.", Num[i][j]); } if (s[i][j]!= ' * ' &&!vis[i][j]) {for (int k=i;s[k][j]!= ' * ' &&am                        p;k<r;k++) {vis[k][j]=1;                    printf ("%c", S[k][j]);                } printf ("\ n"); }}}} return 0;}

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UVA 232-crossword Answers