题目描述

模仿例题，打印出不同方向的正方形，然后打印三角形矩阵。中间有个空行。

输入格式

输入矩阵的规模，不超过 $9$。

输出格式

输出矩形和正方形

输入输出样例 #1

输入 #1

4

输出 #1

01020304
05060708
09101112
13141516

      01
    0203
  040506
07080910

题解

#include<iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int cnt = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (cnt <= 9) cout << 0 << cnt;
            else cout << cnt;
            cnt++;
        }
        cout << endl;
    }
    cout << endl;
    int count = 1;
    for (int a = 1; a <= n; a++)
    {
        int space = 0;
        while (space < 10 - 2 * (a + 1)) 
        {
            cout << " ";
            space++;
        }
        for (int b = 1; b <= a; b++)
        {
            if (count <= 9) cout << 0 << count;
            else cout << count;
            count++;
        }
        cout << endl;
    }
    return 0;
}

但是 WA 了

经过 debug，问题出在空格的输出上。

原本计算空格的公式为10 - 2 * (a + 1)，应该为(n - a) * 2才对。

修改后的代码

#include<iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int cnt = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (cnt <= 9) cout << 0 << cnt;
            else cout << cnt;
            cnt++;
        }
        cout << endl;
    }
    cout << endl;
    int count = 1;
    for (int a = 1; a <= n; a++)
    {
        int space = 0;
        while (space < (n - a) * 2) 
        {
            cout << " ";
            space++;
        }
        for (int b = 1; b <= a; b++)
        {
            if (count <= 9) cout << 0 << count;
            else cout << count;
            count++;
        }
        cout << endl;
    }
    return 0;
}

AC。