zoukankan      html  css  js  c++  java
  • ZOJ Problem Set–1414 Number Steps

    Time Limit: 2 Seconds      Memory Limit: 65536 KB


    Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued.

    You are to write a program that reads the coordinates of a point (x, y), and writes the number (if any) that has been written at that point. (x, y) coordinates in the input are in the range 0...5000.

    Input
    The first line of the input is N, the number of test cases for this problem. In each of the N following lines, there is x, and y representing the coordinates (x, y) of a point.

    Output
    For each point in the input, write the number written at that point or write No Number if there is none.

    Sample Input
    3
    4 2
    6 6
    3 4

    Sample Output
    6
    12
    No Number

    乍一看这题目,我被这图吓到了,难道是什么高深的数学计算吗?仔细一看,原来只是用到了简单的数列知识而已。这个题目结合数列以及直线方程就可以很好的解决了。首先可以意识到这些数字的分布,是在两条直线上的,分别是:

    y = x

    y = x – 2

    而每条直线上的数字又是有两个公差为4的等差数列构成,对于y = x,可以得到数列:

    当x为偶数时:ax = 4*x/2;

    当x为奇数时:ax = 4*(x+1)/2 - 3;

    对于直线y  = x – 2,有:

    当x为偶数时:ax = 4*x/2 – 2

    当x为奇数时:ax = 4*(x – 1)/2 – 1

    有了这些关系之后,程序呼之欲出啊,代码如下:

    #include<iostream>
    
    using namespace std;
    
    int main()
    
    {
    
      /*
    
      y = x
    
      an = 4n - 3 或 an = 4n
    
      y = x - 2
    
      an = 4n - 2 或 an = 4n - 1
    
      */
    
      int paires;cin>>paires;
    
      int x, y;
    
      while(paires-- && cin>>x>>y)
    
      {
    
        if(x == y)
    
        {
    
          if(x%2 == 0)
    
          {
    
            cout<<4*x/2<<endl;
    
          }
    
          else
    
          {
    
            cout<<4*(x+1)/2 - 3<<endl;
    
          }
    
        }
    
        else if( y == x - 2)
    
        {
    
          if(x%2 == 0)
    
          {
    
            cout<<4*x/2 - 2<<endl;
    
          }
    
          else
    
          {
    
            cout<<4*(x - 1)/2 - 1<<endl;
    
          }
    
        }
    
        else
    
          cout<<"No Number"<<endl;
    
      }
    
      return 0;
    
    }
  • 相关阅读:
    Kafka Shell基本命令(包括topic的增删改查)
    thefuck的安装和使用
    Linux运维利器之ClusterShell
    MySQL数据库的10大经典错误案例
    Mysql 常用操作
    Git 忽略特定文件或文件夹
    为什么需要拷贝构造函数
    C语言编译过程
    设计模式之建造者模式
    设计模式之工厂/抽象工厂模式
  • 原文地址:https://www.cnblogs.com/malloc/p/2398307.html
Copyright © 2011-2022 走看看