zoukankan      html  css  js  c++  java
  • cf 118B

    B. Present from Lena
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:


              0
            0 1 0
          0 1 2 1 0
        0 1 2 3 2 1 0
      0 1 2 3 4 3 2 1 0
    0 1 2 3 4 5 4 3 2 1 0
      0 1 2 3 4 3 2 1 0
        0 1 2 3 2 1 0
          0 1 2 1 0
            0 1 0
              0

    Your task is to determine the way the handkerchief will look like by the given n.

    Input

    The first line contains the single integer n (2 ≤ n ≤ 9).

    Output

    Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.

    Sample test(s)
    Input
    2
    Output
        0
    0 1 0
    0 1 2 1 0
    0 1 0
    0
    Input
    3
    Output
          0
    0 1 0
    0 1 2 1 0
    0 1 2 3 2 1 0
    0 1 2 1 0
    0 1 0
    0


     1 #include<map>
     2 #include<cmath>
     3 #include<queue>
     4 #include<cstdio>
     5 #include<vector>
     6 #include<string>
     7 #include<cstring>
     8 #include<sstream>
     9 #include<iostream>
    10 #include<algorithm>
    11 using namespace std;
    12 int main()
    13 {
    14     int n;
    15     cin>>n;
    16     for(int i = 0 ; i < n + 1; i ++)
    17     { 
    18         for(int j = 0 ; j < i * 2 + 1 ; j ++)
    19         {        
    20             for(int k = i*2 ; j==0 && k < n * 2 ; k++ )cout<<" ";
    21             if(j<=i&&j!=i*2)cout<<j<<" ";
    22             else if(j<=i)cout<<j;
    23             else if(j!=i*2)cout<<i*2-j<<" ";
    24                  else cout<<i*2-j;
    25         }
    26         cout<<endl;    
    27     }
    28     for(int i = 0 ; i < n  ; i++)
    29     { 
    30         for(int j = 0 ; j < (n - i )* 2 - 1 ; j++)
    31         {  
    32             for(int k = 0 ; j==0&& k < (i+1)*2 ; k++)cout<<" ";
    33             if(j<=n-i-1 && j!= (n-i)*2-2)cout<<j<<" ";
    34             else if(j<=n-i-1)cout<<j;
    35             else if(j!=(n-i)*2-2)cout<<(n-i -1)*2 -j <<" ";
    36                  else cout<<(n-i-1)*2-j;
    37         }
    38         cout<<endl;
    39     
    40     }
    41 }
  • 相关阅读:
    zookeeper基础
    4. Zookeeper单机版安装
    3. 服务中间件Dubbo
    2. Maven工程的搭建
    Maven的优点
    应用架构的演进历史 MVC、 RPC、SOA 和 微服务架构
    js 正则表达式验证
    后台正则表达式验证部分部分总结
    Linux下记录登录用户历史操作
    Django使用Ace实现在线编辑器
  • 原文地址:https://www.cnblogs.com/Duskcl/p/3973961.html
Copyright © 2011-2022 走看看