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  • 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
", ans);
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5791334.html
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