P5725 【深基4.习8】求三角形
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
输出 #1
1 2 3 4 5 6 7 8 9
| 01020304 05060708 09101112 13141516
01 0203 040506 07080910
|
题解
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| #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
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| #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。