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SDUT 1068-Number Steps（數學：直線）

編輯：C++入門知識

SDUT 1068-Number Steps（數學：直線）

Number Steps

Time Limit: 1000ms Memory limit: 10000K 有疑問？點這裡^_^

題目描述

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.

輸入

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.

輸出

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

示例輸入

示例輸出

6
12
No Number

就是按圖中的規律給出兩條直線。。我一開始居然沒看出來是直線。。找規律打表敲了一大片結果wa了，後來發現就是判斷點是否在直線上嘛 兩條直線分別為y=x與y=x-2; 然後那個編號可以根據坐標x寫出對應關系，很好寫，都是等差數列，我是分奇偶討論的。。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#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\n",2*x-1);
			else
				printf("%d\n",2*x);
		}
		else if(y==x-2)
		{
			if(x%2)
				printf("%d\n",2*x-3);
			else
				printf("%d\n",2*x-2);
		}
		else
			puts("No Number");
	}
    return 0;
}