PKU_ACM1663

Number Steps

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10116 Accepted: 5389

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...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 Input3
4 2
6 6
3 4

 

Sample Output

6
12
No Number

 参考代码如下：

#include "stdio.h"#define NUM 5001int f(int x,int y){int p[4];if(x == 0 && y == 0)return 0;if(x == 1 && y == 1)return 1;if(x >= 2){if(x % 2 == 1){p[0] = 2 * x -3;p[1] = 2 * x -1;}if(x % 2 == 0){p[0] = 2 * x -2;p[1] = 2 * x;}if(y % 2 == 1){p[2] = 2 * y - 1;p[3] = 2 * y + 1;}if(y % 2 == 0){p[2] = 2 * y;p[3] = 2 * y + 2;}if(p[0] == p[2] || p[0] == p[3])return p[0];if(p[1] == p[2] || p[1] == p[3])return p[1];elsereturn -1;}elsereturn -1;}int main(){int x,y,n;scanf("%d",&n);while(n--){scanf("%d %d",&x,&y);if(f(x,y) == -1)printf("No Number/n");elseprintf("%d/n",f(x,y));}return 0;}