PAT basic 1027

来源:互联网 发布:c语言北京电脑学校 编辑:程序博客网 时间:2024/06/07 06:34
#include <iostream>using namespace std;int main() {    int N;    char c;    cin >> N >> c;    int row = 0;    for (int i = 0; i < N; i++) {        if ((2 * i * (i + 2) + 1) > N) {            row = i - 1;            break;        }    }    for (int i = row; i >= 1; i--) {        for (int k = row - i; k >= 1; k--)            cout << " ";        for (int j = i * 2 + 1; j >= 1; j--)            cout << c;        cout << endl;    }    for (int i = 0; i < row; i++)        cout << " ";    cout << c << endl;    for (int i = 1; i <= row; i++) {        for (int k = row - i; k >= 1; k--)            cout << " ";        for (int j = i * 2 + 1; j >= 1; j--)            cout << c;        cout << endl;    }    cout << (N - (2 * row * (row + 2) + 1));    return 0;}
原创粉丝点击