PAT basic 1036

来源:互联网 发布:软件成本估算方法 编辑:程序博客网 时间:2024/06/03 21:54
#include <iostream>using namespace std;int main() {    int N;    char c;    cin >> N >> c;    int t = N / 2 + N % 2;    for (int i = 0; i < N; i++)        cout << c;    cout << endl;    for (int i = 0; i < t - 2; i++) {        cout << c;        for (int k = 0; k < N - 2; k++)            cout << " ";        cout << c << endl;    }    for (int i = 0; i < N; i++)        cout << c;    return 0;}
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