2014北大软工夏令营机试 B:Number Steps_北大软微夏令营手写代码-优快云博客

本文链接：https://blog.youkuaiyun.com/moweiyang0214/article/details/37973027

本文介绍了一个从平面坐标系中的点映射到特定数列的问题，通过给出的坐标(x, y)，程序将返回该坐标对应的数列中的数字。如果坐标不在定义的数列模式中，则输出'NoNumber'。此问题适用于算法训练，特别是路径查找和模式识别方面。

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B:Number Steps

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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

样例输入

样例输出

6
12
No Number

#include<iostream>
using namespace std;

int Query(int a,int b){
   	int m=0,n=0,sum=0;
   	while(true){
   		if(m==a&&n==b)
		   break;
		else{
			sum++;
			int b=sum%4;
			switch(b){
			    case 0:
			    	m--;
			    	n++;
				    break;
				 case 1:
				 case 3:
				 	m++;
				 	n++;
				 	break;
				 case 2:
				 	m++;
				 	n--;
				 	break;
				 
			}	
		}
   	}
   	return sum;
}


int main(){
	int n,x,y;
	cin>>n;
	for(int i=0;i<n;i++){
       cin>>x>>y;
	   if(y==x||y==(x-2))	
	      cout<<Query(x,y)<<endl;
	    else{
	    	cout<<"No Number"<<endl;	
	    }
	}
	return 0;
}