HDOJ-1391

Number Steps

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4987    Accepted Submission(s): 3030

Problem 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 Input

3

4 2

6 6

3 4

 

 

Sample Output

6

12

No Number

 

题意：如上图的一个坐标图，给出x,y，输出对应点的值，如果点为空则输出No Number。

找规律，我们可以发现x和y只有相等和x-2=y两种情况，且当x为偶数时点的值为x+y，为奇数时为x+y-1。

AC代码：

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

int main(){
    int n,x,y;
    while(~scanf("%d",&n)){
        while(n--){
            scanf("%d %d",&x,&y);
            if(x==y||x-2==y){
                printf("%d
",x%2==0?x+y:x+y-1);//当x%2==0成立时，进行x+y；否则进行x+y-1 
            }
            else
            printf("No Number
");
        }
    }
    return 0;
}