D.Cypher Game

Cypher Game

n*m的矩阵只能向右读数，相当于一层又一层的 嵌套，然后根据读取的顺序，输入两行等长的数字1-n*m,第一行和和第二行是一一对应的，按照读取的顺序输出第二行的解密密码。

C++

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
int a[maxn][maxn];
int b[maxn * maxn + 10];
int c[maxn * maxn + 10];
int cnt = 1;
void work(int n, int m)
{

    if (n == 1)
    {
        for (int i = 1; i <= m; i++)
            cout<<b[a[1][i]];
        return ;
    }
    else if (m == 1)
    {
        for (int i = 1; i <= n; i++)
            cout<<b[a[i][1]];
        return ;
    }
    else
    {
        int i = 1, j = 1;
        int sum = n * m;
        while (sum>0)
        {
            for (int k = 1; k < m; k++)
            {
                sum--;
                cout<<b[a[i][j]];
                j++;
            }
            for (int k = 1; k < n; k++)
            {
                sum--;
                cout<<b[a[i][j]];
                i++;
            }
            for (int k = 1; k < m; k++)
            {
                sum--;
                cout<<b[a[i][j]];
                j--;
            }
            for (int k = 1; k < n; k++)
            {
                sum--;
                cout<<b[a[i][j]];
                i--;
            }
            i++;
            j++;
            n -= 2;
            m -= 2;
            if (n == 1)
            {
                for (int k = 1; k <= m; k++)
                {
                    sum--;
                    cout<<b[a[i][j]];
                    j++;
                }
            }
            else if (m == 1)
            {
                for (int k = 1; k <= n; k++)
                {
                    sum--;
                    cout<<b[a[i][j]];
                    i++;
                }
            }
        }
    }
}
int main()
{
    // freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    for (int i = 1; i <=n * m; i++)
        cin >> c[i];
    for (int i = 1; i <= n * m; i++)
        cin>>b[c[i]];
    work(n, m);
    return 0;
}

C

#include<stdio.h>
int a[1001][1001], b[1001 * 1001], c[1001 * 1001], A, B, C;
int main()
{
    int n, m, i, j;
    scanf("%d%d", &n, &m);
    for (i = 0; i < n; i++)
        for (j = 0; j < m; j++)
            scanf("%d", &a[i][j]);
    for (i = 0; i < n * m; i++)
        scanf("%d", &b[i]);
    for (i = 0; i < n * m; i++)
        scanf("%d", &c[b[i] - 1]);
    for (A = 0, B = n - 1, C = m - 1; A <= B && A <= C; A++, B--, C--)
    {
        if (A == B)
        {
            for (i = A, j = A; j <= C; j++)
                printf("%d", c[a[i][j] - 1]);
            continue;
        }
        if (A == C)
        {
            for (i = A, j = A; i <= B; i++)
                printf("%d", c[a[i][j] - 1]);
            continue;
        }
        for (i = A, j = A; j <= C; j++)
            printf("%d", c[a[i][j] - 1]);
        for (i = A + 1, j = C; i <= B - 1; i++)
            printf("%d", c[a[i][j] - 1]);
        for (i = B, j = C; j >= A; j--)
            printf("%d", c[a[i][j] - 1]);
        for (i = B - 1, j = A; i >= A + 1; i--)
            printf("%d", c[a[i][j] - 1]);
    }
    return 0;
}