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最新推荐文章于 2020-03-29 22:19:55 发布

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最新推荐文章于 2020-03-29 22:19:55 发布

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原文链接：http://www.cnblogs.com/RootVount/p/10990690.html

本文探讨了从平面坐标系原点出发，将所有非负整数按特定模式分布在平面上的数学问题。通过分析，给出了一种算法，能够根据输入的坐标(x,y)，判断该点上是否写有特定的非负整数，并输出该数值或NoNumber。示例展示了如何处理坐标范围为0到1000000的情况。

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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...1000000.

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

Hint

水

Source

Unknown

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long

int n, x, y;

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 0; i<n; i++)
        {
            scanf("%d%d", &x, &y);
            if((x+y)%2 == 0)
            {
                if((x == y || x-y == 2) && x%2 == 1)printf("%d\n", x+y-1);
                else if((x == y || x-y == 2) && x%2 == 0)printf("%d\n", x+y);
                else if(x != y || x-y != 2) printf("No Number\n");
            }
            else
                printf("No Number\n");
        }
    }
    return 0;
}