Number Steps

Number Steps

时间限制(普通/Java):3000MS/10000MS          运行内存限制:65536KByte

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

3

4 2

6 6

3 4

 

样例输出

6

12

No Number

 

题解：由图可以看出此图由两条直线组成

     第一条：y = x

     第二条：y = x - 2

     且每条直线上的数字由两组公差为4的数列组成，如0,4,8,12 ； 1,5,9,13.

     故若y=x,当x为偶数时，输出2*x;当x为奇数时输出2*x-1;

     若y=x-2,当x为偶数时，输出2*x-2;当x为奇数时输出2*x-3.

 

 

 

 

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int main()
{
    int TT;
    int x,y;
    scanf("%d",&TT);
    while(TT--)
    {
        scanf("%d%d",&x,&y);
        if(x==y)
        {
            if(x%2==0)
            printf("%d\n",2*x);
            else
            printf("%d\n",2*x-1);
        }
        else if(x-y==2)
        {
            if(x%2==0)
            printf("%d\n",2*x-2);
            else
            printf("%d\n",2*x-3);
        }
        else
        printf("No Number\n");
    }
    return 0;
}