數字階梯（Number Steps）

Number Steps

Time Limit: 1 Second      Memory Limit: 32768 KB
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 

解答：View Code

int main()
{
int N, i;
    scanf("%d", &N);   //N表示輸入檔案中測試資料的數目
    int x, y;
for (i = 0; i < N; i++)   //處理輸入檔案中的N個測試資料
    {
        scanf("%d %d", &x, &y);   //每個測試資料包含兩個整數
        int Num;                  //在(x,y)座標點上的非負整數
        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;
}