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    After the 32nd ACM/ICPC regional contest, Wiskey is beginning to prepare for CET-6. He has an English words table and read it every morning.
    One day, Wiskey's chum wants to play a joke on him. He rolling the table, and tell Wiskey how many time he rotated. Rotate 90 degrees clockwise or count-clockwise each time.
    The table has n*n grids. Your task is tell Wiskey the final status of the table.

    InputEach line will contain two number.
    The first is postive integer n (0 < n <= 10).
    The seconed is signed 32-bit integer m.
    if m is postive, it represent rotate clockwise m times, else it represent rotate count-clockwise -m times.
    Following n lines. Every line contain n characters.
    OutputOutput the n*n grids of the final status.
    Sample Input

    3 2
    123
    456
    789
    3 -1
    123
    456
    789

    Sample Output

    987
    654
    321
    369
    258
    147

    把数据放在矩阵里然后直接改变输出方向就可以了、

    #include<iostream>
    using namespace std;
    int main()
    {
    char num[11][11];
    int n, m;
    while (cin >> n >> m)
    {
    for (int i = 1; i <= n;i++)
    for (int j = 1; j <= n; j++) cin >> num[i][j];
    m %= 4;
    if (m < 0) m = 4 + m;
    if (m == 0)
    {
    for (int i = 1; i <= n; i++)
    {
    for (int j = 1; j <= n; j++)
    {
    cout << num[i][j];
    }
    cout << endl;
    }
    }
    else if (m == 1)
    {
    for (int i = 1; i <= n; i++)
    {
    for (int j = n; j >= 1; j--)
    {
    cout << num[j][i];
    }
    cout << endl;
    }
    }
    else if (m == 2)
    {
    for (int i = n; i >= 1; i--)
    {
    for (int j = n; j >= 1; j--)
    {
    cout << num[i][j];
    }
    cout << endl;
    }
    }
    else if (m == 3)
    {
    for (int i = n; i >= 1; i--)
    {
    for (int j = 1; j <= n; j++)
    {
    cout << num[j][i];
    }
    cout << endl;
    }
    }
    }
    return 0;
    }


    我觉得这个题可以练习一下找bug的技巧。将它的转几次改为把每一次都输出,来寻找bug更好
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  • 原文地址:https://www.cnblogs.com/damaoranran/p/8778270.html
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