UVALive 5886 The Grille （模拟）

The Grille

题目链接：

http://acm.hust.edu.cn/vjudge/problem/26634

Description

http://7xjob4.com1.z0.glb.clouddn.com/a7d33cb3303ea18c9e6f3de676f242a6

Input

The input contains several test cases. Each test case contains description of a grille and a ciphertext.
Your task is to decipher the message and write the plaintext to output.
Each test case starts with a line containing number N (1 ≤ N ≤ 1000), where N is the size of the
grille. Then there are N lines containing the grille description. Each of those lines contains exactly N
characters which are either the “hash” character ‘#’ (solid/opaque material) or the uppercase letter ‘O’
(hole).
Note: In praxis, the grille holes would be arranged in such a way that no position of the ciphertext
is used more than once. In our problem, this is not guaranteed. Some grilles may contain holes that
match the same position/letter of the ciphertext (after rotations). However, the deciphering algorithm
is still the same.
After the grille description, there are another N lines with the enciphered message. Each of them
contains exactly N characters - uppercase letters of alphabet.
The last test case is followed by a line containing one zero.

Output

For each test case, output the deciphered message (plaintext) on one line with no spaces.

Sample Input

```
4
##O#
#O#O
####
###O
ARAO
PCEM
LEEN
TURC
3
O#O
###
O#O
ABC
DEF
GHI
0
```

Sample Output

ACMCENTRALEUROPE
ACGIACGIACGIACGI

Source

2016-HUST-线下组队赛-1


##题意：

加密过程：每次在空白位置写一个字符，写完后正旋90°再写，重复四次.
求解密后的字符串.


##题解：

瞎转一通就好了.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 1010
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

char mes[maxn][maxn];

char key[maxn][maxn];

char ans[maxnmaxn4];

int main(int argc, char const *argv[])

{

//IN;

while(scanf("%d", &n) != EOF && n)
{
    for(int i=1; i<=n; i++) {
        scanf("%s", key[i]+1);
    }
    for(int i=1; i<=n; i++) {
        scanf("%s", mes[i]+1);
    }

    int cnt  = 0;
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) if(key[i][j] == 'O') {
            ans[cnt++] = mes[i][j];
        }
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) if(key[n-j+1][i] == 'O') {
            ans[cnt++] = mes[i][j];
        }
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) if(key[n-i+1][n-j+1] == 'O') {
            ans[cnt++] = mes[i][j];
        }
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<=n; j++) if(key[j][n-i+1] == 'O') {
            ans[cnt++] = mes[i][j];
        }
    }

    ans[cnt] = 0;
    printf("%s\n", ans);
}

return 0;

}