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
題意：如上圖的一個坐标圖，給出x,y，輸出對應點的值，如果點為空則輸出No Number。
找規律，我們可以發現x和y隻有相等和x-2=y兩種情況，且當x為偶數時點的值為x+y，為奇數時為x+y-1。
AC代碼：
#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);//當x%2==0成立時，進行x+y；否則進行x+y-1
}
else
printf("No Number
");
}
}
return 0;
}