数字阶梯（Number Steps）

Number Steps

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

 

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解答：

int main()
{
    int N, i;
    scanf("%d", &N);   //N表示输入文件中测试数据的数目
    int x, y;
    for (i = 0; i < N; i++)   //处理输入文件中的N个测试数据
    {
        scanf("%d %d", &x, &y);   //每个测试数据包含两个整数
        int Num;                  //在(x,y)坐标点上的非负整数
        if (y != x && y != x-2)
        {
            Num = -1;
        }
        else
        {
            if (y == x && x % 2 == 0)
            {
                Num = 2 * x;
            }
            else if (y == x && x % 2 != 0)
            {
                Num = 2 * x - 1;
            }
            else if (y == x-2 && x%2 == 0)
            {
                Num = 2 * x -2;
            }
            else
            {
                Num = 2 * x -3;
            }
        }
        if (Num == -1) printf("No Number\n");
        else printf("%d\n", Num);
    }
    return 1;
}