//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define inf 0x3f3f3f3f
#define eps 1e-6
using namespace std;
#define clr(x) memset(x,0,sizeof((x)))
const int maxn = 5e4+1;//2e6+1
#define MAX(a,b,c) ((a)>(b)?((a)>(c)?(a):(c)):((b)>(c)?(b):(c)))
#define _max(a,b) ((a) > (b) ? (a) : (b))
#define _min(a,b) ((a) < (b) ? (a) : (b))
#define _for(a,b,c) for(int a = b;a<c;a++)
char a[15][15],b[15][15],c[15][15],d[15][15];
char tmp[15][15];
int n,flag = 1;
void fun(char x[15][15],char y[15][15]) {
_for(i,0,n) {
_for(j,0,n) {
x[j][n-i-1] = y[i][j];
}
}
_for(i,0,n) {
_for(j,0,n) {
if(x[i][j]!=b[i][j]) {
flag = 0;
break;
}
}
}
}
void assignment(char x[15][15],char y[15][15]) {
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
x[i][j] = y[i][j];
}
}
}
//反射
void reflex() {
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
c[i][n-j-1] = a[i][j];
}
}
}
void refjud() {
_for(i,0,n) {
_for(j,0,n) {
if(c[i][j]!=b[i][j]) {
flag = 0;
break;
}
}
}
}
int main() {
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
cin>>n;
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
cin>>a[i][j];
}
}
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
cin>>b[i][j];
}
}
assignment(tmp,a);
for(int i = 0;i<3;i++) {
fun(c,tmp);
if(flag){
cout<<i+1;
return 0;
}
flag = 1;
assignment(tmp,c);
clr(c);
}
reflex();
refjud();
if(flag){
cout<<4;
return 0;
}
flag = 1;
for(int i = 0;i<3;i++) {
fun(d,c);
if(flag) {
cout<<5;
return 0;
}
flag = 1;
assignment(c,d);
clr(d);
}
_for(i,0,n) {
_for(j,0,n) {
if(a[i][j]!=b[i][j]) {
flag = 0;
break;
}
}
}
if(flag){
cout<<6;return 0;}
cout<<7;
return 0;
}
洛谷 P1205 [USACO1.2]方块转换 Transformations C/C++ 模拟
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转载自blog.csdn.net/Jason__Jie/article/details/112431343
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