SDUT 1068-Number Steps（数学：直线）

Number Steps

Time Limit: 1000ms   Memory limit: 10000K  有疑问？点这里^_^

题目描写叙述

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.

演示样例输入

3
4 2
6 6
3 4

演示样例输出

6
12
No Number

就是按图中的规律给出两条直线。。
我一開始竟然没看出来是直线。。找规律打表敲了一大片结果wa了，后来发现就是推断点是否在直线上嘛 两条直线分别为y=x与y=x-2; 然后那个编号能够依据坐标x写出相应关系，非常好写，都是等差数列。我是分奇偶讨论的。。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define ll long long
using namespace std;
const int INF=1<<27;
const int maxn=1010;
int main()
{
	int x,y,n;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&x,&y);
		if(x==y)
		{
			if(x%2)
				printf("%d
",2*x-1);
			else
				printf("%d
",2*x);
		}
		else if(y==x-2)
		{
			if(x%2)
				printf("%d
",2*x-3);
			else
				printf("%d
",2*x-2);
		}
		else
			puts("No Number");
	}
    return 0;
}