poj 1663 Number Steps

本文介绍了一个基于平面坐标系的特殊模式,该模式将所有非负整数映射到坐标点上。通过输入坐标(x,y),程序可以返回对应位置上的整数。文章提供了具体的实现代码,并展示了如何通过条件判断来确定坐标点上是否存在数字及其具体数值。

Number Steps

Time Limit: 1000MS
Memory Limit: 10000K

Total Submissions: 11605
Accepted: 6167

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

3
4 2
6 6
3 4

Sample Output

6
12
No Number
  1: #include<iostream>
  2: using namespace std;
  3: int main()
  4: {
  5: 	int n;
  6: 	int x,y;
  7: 	cin>>n;
  8: 	while(n--)
  9: 	{
 10: 		cin>>x>>y;
 11: 		if(x-y!=0 && x-y!=2)
 12: 			cout<<"No Number"<<endl;
 13: 		else if(y%2==0 && x-y==0)
 14: 			cout<<2*y<<endl;
 15: 		else if(y%2==0 && x-y==2)
 16: 			cout<<2*y+2<<endl;
 17: 		else if(y%2==1 && x-y==0)
 18: 			cout<<2*y-1<<endl;
 19: 		else if(y%2==1 && x-y==2)
 20: 			cout<<2*y+1<<endl;
 21: 	}
 22: 	return 0;
 23: }
 24: 

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2257668.html

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