POJ 1663：Number Steps

最新推荐文章于 2020-09-09 11:06:46 发布

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最新推荐文章于 2020-09-09 11:06:46 发布

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分类专栏：水题 POJ 文章标签： POJ

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本文介绍了一种在特定平面坐标系中查找对应整数的算法。该算法通过判断输入坐标(x, y)的位置来确定是否能映射到一个非负整数上，并详细展示了如何根据坐标位置计算出具体数值的方法。

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Number Steps

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13664		Accepted: 7378

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.

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

Sample Output

6
12
No Number

洪水题，找规律。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	int Test,x,y;
	cin>>Test;
	
	while(Test--)
	{
		cin>>x>>y;
		if((x==y||x-2==y)&&x>=0&&y>=0)
		{
			if(x==y)
			{
			    if(x%2)
				{
					cout<<x*2-1<<endl;
				}
				else
				{
					cout<<x*2<<endl;
				}
			}
			else
			{
				if(x%2)
				{
					cout<<x*2-3<<endl;
				}
				else
				{
					cout<<x*2-2<<endl;
				}
			}
		}
		else
		{
			cout<<"No Number"<<endl;
		}
	}
	return 0;
}