1. 被这道题虐了无数次，每次编译通过，但是运行会错误（当时用的是string数组）， 后来把string数组改成char[][]二维数组就好了；
2. 给的参考代码是不对的
3. 如果函数的返回类型是数组，只能通过指针来实现，但是参考代码给了一种巧妙的方法，把数组封装成结构体，然后返回结构体
以下是我的代码：
/*
ID: dollar4
PROG: transform
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>
#include <cstring>
using namespace std;
typedef char array[10][10];
char ori[10][10], tasf[10][10];
char rst[10][10];
int n;
int rot90(char arr[][10], array &cag)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
{
cag[j][n-1-i] = arr[i][j];
}
return 1;
}
int rot180(char arr[][10], array &cag)
{
char cag1[10][10];
rot90(arr, cag1);
rot90(cag1, cag);
return 1;
}
int rot270(char arr[][10], array &cag)
{
char cag1[10][10];
rot180(arr, cag1);
rot90(arr, cag);
return 1;
}
int reflect(char arr[][10], array &cag)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
cag[i][n-j-1] = arr[i][j];
return 1;
}
int grp90(char arr[][10], array &cag)
{
char cag1[10][10];
reflect(arr, cag1);
rot90(cag1, cag);
return 1;
}
int grp2(char arr[][10], array &cag)
{
char cag1[10][10];
reflect(arr, cag1);
rot180(cag1, cag);
return 1;
}
int grp3(char arr[][10], array &cag)
{
char cag1[10][10];
reflect(arr, cag1);
rot270(cag1, cag);
return 1;
}
bool equ(char arr1[][10], char arr2[][10])
{
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (arr1[i][j] != arr2[i][j])
return false;
else continue;
return true;
}
int main()
{
ofstream fout ("transform.out");
ifstream fin ("transform.in");
int i, j;
fin >> n;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
fin >> ori[i][j];
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
fin >> tasf[i][j];
if (rot90(ori, rst) && equ(rst, tasf))
fout << 1 << endl;
else if (rot180(ori, rst) && equ(rst, tasf))
fout << 2 << endl;
else if (rot270(ori, rst) && equ(rst, tasf))
fout << 3 << endl;
else if (reflect(ori, rst) && equ(rst, tasf))
fout << 4 << endl;
else if (grp90(ori, rst) && equ(rst, tasf))
fout << 5 << endl;
else if (grp2(ori, rst) && equ(rst, tasf))
fout << 5 << endl;
else if (grp3(ori, rst) && equ(rst, tasf))
fout << 5 << endl;
else if (equ(ori, tasf))
fout << 6 << endl;
else fout << 7 << endl;
return 0;
}
以下是参考代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAXN 10
typedef struct Board Board;
struct Board {
int n;
char b[MAXN][MAXN];
};
/* rotate 90 degree clockwise: [r, c] -> [c, n+1 - r] */
Board
rotate(Board b)
{
Board nb;
int r, c;
nb = b;
for(r=0; r<b.n; r++)
for(c=0; c<b.n; c++)
nb.b[c][b.n+1 - r] = b.b[r][c];
return nb;
}
/* reflect board horizontally: [r, c] -> [r, n-1 -c] */
Board
reflect(Board b)
{
Board nb;
int r, c;
nb = b;
for(r=0; r<b.n; r++)
for(c=0; c<b.n; c++)
nb.b[r][b.n-1 - c] = b.b[r][c];
return nb;
}
/* return non-zero if and only if boards are equal */
int
eqboard(Board b, Board bb)
{
int r, c;
if(b.n != bb.n)
return 0;
for(r=0; r<b.n; r++)
for(c=0; c<b.n; c++)
if(b.b[r][c] != bb.b[r][c])
return 0;
return 1;
}
Board
rdboard(FILE *fin, int n)
{
Board b;
int r, c;
b.n = n;
for(r=0; r<n; r++) {
for(c=0; c<n; c++)
b.b[r][c] = getc(fin);
assert(getc(fin) == '\n');
}
return b;
}
void
main(void)
{
FILE *fin, *fout;
Board b, nb;
int n, change;
fin = fopen("transform.in", "r");
fout = fopen("transform.out", "w");
assert(fin != NULL && fout != NULL);
fscanf(fin, "%d\n", &n);
b = rdboard(fin, n);
nb = rdboard(fin, n);
if(eqboard(nb, rotate(b)))
change = 1;
else if(eqboard(nb, rotate(rotate(b))))
change = 2;
else if(eqboard(nb, rotate(rotate(rotate(b)))))
change = 3;
else if(eqboard(nb, reflect(b)))
change = 4;
else if(eqboard(nb, rotate(reflect(b)))
|| eqboard(nb, rotate(rotate(reflect(b))))
|| eqboard(nb, rotate(rotate(rotate(reflect(b))))))
change = 5;
else if(eqboard(nb, b))
change = 6;
else
change = 7;
fprintf(fout, "%d\n", change);
exit(0);
}