#include <iostream> using namespace std; int main() { int n; // 读取输入的矩阵行数和列数 cin >> n; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (j == 0 || j == n - 1) { // 每行的第一个和最后一个位置输出金色小行星 cout << '#'; } else { // 其他位置输出普通小行星 cout << '*'; } } // 每行输出结束后换行 cout << endl; } return 0; } 去除备注
#include <iostream> using namespace std; int main() { int n; cin >> n; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (j == 0 || j == n - 1) { cout << '#'; } else { cout << '*'; } } cout << endl; } return 0; } [2026-01-31 19:59:37 | AI问答 | 116点数解答]
相关提问
- [字节豆包] 对于一元二次方程 $$ax^2 + bx + c = 0,解可以分为很多情况。若该方程有两个不相等实根，首先输出1，换行，然后从小到大输出两个实根，换行；若该方程有两个相等实根，首先输出2，换行，然后输出这个这个实根，换行；若该方程有一对共轭复根，输出3，换行；若该方程无解，输出4，换行；若该方程有无穷个解，输出5，换行；若该方程只有一个根，首先输出6，换行，然后输出这个跟，换行；要求使用 c++ class 编写程序。可以创建如下 class #include <iostream> #include <cmath> using namespace std; class Equation{ private: int _a, _b, _c; public: Equation(int a, int b, int c){ } void solve(){ } }; int main(){ int a, b, c; cin >> a >> b >> c; Equation tmp(a, b, c); tmp.solve()（481点数解答 | 2025-03-07 12:49:26）180
- [字节豆包] ```cpp #include <iostream> using namespace std; int main() { int n; cin >> n; // 输入年龄 if (n >= 0 && n <= 3) { cout << "infant" << endl; } else if (n >= 4 && n <= 12) { cout << "child" << endl; } else if (n >= 13 && n <= 18) { cout << "youngster" << endl; } else if (n >= 19 && n <= 25) { cout << "youth" << endl; } return 0; }（780点数解答 | 2026-01-29 16:57:33）6
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- [阿里通义] 快速排序：# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"（951点数解答 | 2024-11-14 10:46:29）314
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