HDU 1391 Number Steps

HDU 1391 Number Steps解析

最新推荐文章于 2019-05-18 17:59:45 发布

原创最新推荐文章于 2019-05-18 17:59:45 发布 · 220 阅读

0 ·

CC 4.0 BY-SA版权

c语言专栏收录该内容

67 篇文章

订阅专栏

本文解析了HDU 1391 Number Steps题目，介绍了一个基于坐标寻找对应数值的方法。通过分析坐标规律，实现了输入坐标(x, y)后返回该位置上的整数，如果该位置未分配整数则输出No Number。

http://acm.hdu.edu.cn/showproblem.php?pid=1391

Number Steps

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2014 Accepted Submission(s): 1277

Problem Description

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.

HDU <wbr>1391 <wbr>Number <wbr>Steps

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
  

Source

Asia 2000, Tehran (Iran)

Recommend

Ignatius.L

题意：给出坐标，求对应坐标的值。

分析：看图找规律。

代码如下:

#include<iostream>
using namespace std;
int main()
{
int T,x,y;
scanf("%d",&T);
while(T--)
{
  scanf("%d %d",&x,&y);
  if(x!=y && y+2!=x)
  {
   printf("No Number\n");
  }
  else if(x==y)
  {
   if(x%2==0)
    printf("%d\n",x+y);
   else
    printf("%d\n",x+y-1);
  }
  else
  {
   if(x%2==0)
    printf("%d\n",2*x-2);
   else
    printf("%d\n",2*x-3);
  }
}
return 0;
}