Poj Solution 3106

http://poj.org/problem?id=3106

#include <stdio.h>


static char str[100010];
static char input[310][310], ans[310][310];
static int n, m;

static int doit( int s ) {
    int i = s/1000, j = s%1000, a = i, b = j;
    int nn = n;
    int mm = m, t;
    
    for( int k=0; str[k]; k++ ) {
        switch( str[k] ){
        case '1':
            a = j; b = i;
            t = nn; nn = mm; mm = t; 
            break;
        case '2':
            a = nn-j; b = mm-i; 
            t = nn; nn = mm; mm = t; 
            break;
        case 'H':
            a = mm-i; b = j; break;
        case 'V':
            a = i; b = nn-j; break;
        case 'B': case 'Y':
            a = mm-i; b = nn-j; break;
        case 'A': case 'Z':
            a = j; b = mm-i;
            t = nn; nn = mm; mm = t; 
            break;
        case 'C': case 'X':
            a = nn-j; b = i; 
            t = nn; nn = mm; mm = t; 
            break;
        }
        i = a;
        j = b;
    }
    return i*1000+j;
}

int main( ){
    int i, j;
    scanf( "%d%d", &m, &n );

    for( i=0; i<m; i++ )
        scanf( "%s", input[i] );

    getchar( );
    gets( str );
    
    int *x=new int[3], *y = new int[3];
    int t;
    n--; m--;
    t = doit( 0*1000+0 );
    x[0] = t/1000; y[0] = t%1000;
    
    t = doit( m*1000+0 );
    x[1] = t/1000; y[1] = t%1000;
    
    t = doit( 0*1000+n );
    x[2] = t/1000; y[2] = t%1000;
    n++;m++;

    int a = x[0], b = y[0], aa, bb;
    int da = ((x[1]==x[0]) ? 0 : ((x[1]>x[0])?1:-1) ),
        db = ((y[1]==y[0]) ? 0 : ((y[1]>y[0])?1:-1) ),
        da1= ((x[2]==x[0]) ? 0 : ((x[2]>x[0])?1:-1) ),
        db1= ((y[2]==y[0]) ? 0 : ((y[2]>y[0])?1:-1) );

    int nn = 0, mm = 0;

    for( i=0; i<m; i++ ) {
        aa = a; bb = b;
        for( j=0; j<n; j++ ) {
            ans[aa][bb] = input[i][j];
            if( aa > nn )
                nn = aa;
            if( bb > mm )
                mm = bb;
            aa += da1; bb += db1;
        }
        a += da; b += db;
    }
    nn++;mm++;
    
    printf( "%d %dn", nn, mm );
    
    for( i=0; i<nn; i++ ) {
        for( j=0; j<mm; j++ )
            printf( "%c", ans[i][j] );
        printf( "n" );
    }
            
    return 0;
}