C Number Steps（九度OJ 1136）

题目描述：

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.

输入：

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.

输出：

For each point in the input, write the number written at that point or write No Number if there is none.

样例输入：

34 26 63 4

样例输出：

612No Number

来源：

2008年北京大学软件所计算机研究生机试真题

题目分析：

       根据输入的点坐标（x，y）输出坐标对应的数，通过观察上图可发现：

       （1）当x=y时，若x、y为偶数，则对应的数为x+y；若x、y为奇数，则对应的数为x+y-1；

       （2）当x！=y，且x-y=2时，若x、y为偶数，则对应的数为x+y；若x、y为奇数，则对应的数为x+y-1；

       （3）当x=y=0时，对应的数为0；

       （4）其它情况，输出No Number，对应的数不存在。

源代码：

#include <stdio.h>int num(int x,int y)                              //返回计算结果，如果输出“No Number”则返回-1{    if(x==0 && y==0)        return 0;    else if((x==y||x-y==2) && x%2==1 && y%2==1)        return x+y-1;    else if((x==y||x-y==2) && x%2==0 && y%2==0)        return x+y;    else        return -1;}int main(){    int N;    int x,y;    int result;    scanf("%d",&N);    while(N>0)    {        scanf("%d %d",&x,&y);        result=num(x,y);        if(result!=-1)            printf("%d\n",result);        else            printf("No Number\n");        N--;    }    return 0;}

程序截图：