Printing numbers in a pyramid pattern（一）

最新推荐文章于 2020-04-29 23:33:56 发布

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本文深入探讨了程序设计中的核心技巧与算法实现方法，包括数据结构、算法设计、版本控制、项目管理等多个方面，旨在帮助开发者提升编程能力与效率。

Description
Write a program that reads an interger n, then display one of the following patterns.
Pattern 1
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6

Pattern 2
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1

Pattern 3
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6

Pattern 4
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1

Input
An integer n (1<=n<=4).

Output
Pattern n.
You should specify the width of each number’s print field to 3, justify the output to right.

Sample Input
Copy sample input to clipboard
1
Sample Output
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6

代码实现:

#include<iostream>
using namespace std;

void  fun1(int rows)
{
    for(int i = 1; i <= rows; i++){
        for(int j = 1; j <= i; j++){
            cout << "  " << j;   //两个空格
        }
        cout << endl;
    }
}

void fun2(int rows)
{
    for(int i = rows; i >= 1; i--){
        for(int j = 1; j <= i; j++){
            cout << "  " << j;    //两个空格
        }
        cout << endl;
    }
}

void fun3(int rows)
{
    for(int i = rows; i >= 1; i--){
        for(int j = 0; j < i-1; j++){
            cout << "   ";  //三个空格 
        }
        for(int j = 1; j <= (rows+1)-i; j++){
            cout << "  " << j;  //两个空格 
        }
        cout << endl;
    }
}

void fun4(int rows)
{
    for(int i = 1; i <= rows; i++){
        for(int j = 0; j < i-1; j++){
            cout << "   ";
        }
        for(int j = 1; j <= (rows+1)-i; j++){
            cout << "  " << j;
        }
        cout << endl;
    }   
} 


int main()
{
    int n;
    while(cin >> n){
        switch(n){
            case 1: fun1(6); break;
            case 2: fun2(6); break;
            case 3: fun3(6); break;
            case 4: fun4(6); break; 
        }
    }
    return 0;
}