这道题要求找到一种操作方式对字符串数组a进行操作,使得其与b一致。一共有六种操作方式,
首先定义三个数组,a,b,c,前两个数组是输入,然后c数组每次选择操作方式之前使其等于a,
然后对c进行操作,这道题值得注意的是操作选择是从1到6,必须按照顺序来,进行每一种操作
后, 将c与b做比较,如果一致,就输出当前操作的编号,如果找不到操作方式,那么就输出 7.
/*
ID:yucept21
LANG:C++
TASK:transform
*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
const int N = 15;
char a[N][N], b[N][N], c[N][N];
int n;
int idea( char a[N][N], int cur)
{
char b[N][N];
memcpy( b, a, sizeof b);
for( int i = 1; i <= n; i ++)
for( int j = 1; j <= n; j ++)
{
if( cur == 1) a[i][j] = b[n - j + 1][i]; //顺时针90°
if( cur == 2) a[i][j] = b[n - i + 1][ n - j + 1]; //顺时针180°
if( cur == 3) a[i][j] = b[j][ n - i + 1]; //顺时针270°
if( cur == 4) a[i][j] = b[i][ n - j + 1]; //水平翻转
}
}
int get()
{
for( int i = 1; i <= n; i ++)
for( int j = 1; j <= n; j ++)
c[i][j] = a[i][j];
}
bool check()
{
for( int i = 1; i <= n; i ++)
for( int j = 1; j <= n; j ++)
if( c[i][j] != b[i][j])
return false;
return true;
}
void make( char a[N][N])
{
char s[15];
for( int i = 1; i <= n; i ++)
{
scanf( "%s", s);
for( int j = 1; j <= n; j ++)
a[i][j] = s[j - 1];
}
}
int main()
{
freopen( "transform.in", "r", stdin);
freopen( "transform.out", "w", stdout);
scanf( "%d", &n);
make( a );
make( b );
for( int i = 1; i <= 4; i ++)
{
get();
idea( c, i);
if( check() ) {
printf( "%d\n", i);
return 0;
}
}
for( int i = 1; i <= 3; i ++)
{
get();
idea( c, 4);
idea( c, i);
if( check() ) {
printf( "5\n");
return 0;
}
}
get();
if( check() )
{
printf( "6\n");
return 0;
}
printf( "7\n");
return 0;
}