Light OJ - 1008 - Fibsieve`s Fantabulous Birthday 题解

最新推荐文章于 2020-11-20 17:35:14 发布

最新推荐文章于 2020-11-20 17:35:14 发布 · 188 阅读

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#数据结构与算法

本文介绍了一种计算特殊长方形中特定时刻点亮的单元格坐标的方法。通过等差数列的推导，给出了一种高效的算法实现，避免了使用循环带来的效率问题。

就是计算特殊长方形坐标的问题，这类题目需要的是细心，细心，再细心，然后推导摸索出公式来，

题目如下：

Fibsieve had a fantabulous (yes, it's an actual word) birthday party this year. He had so many gifts that he was actually thinking of not having a party next year.

Among these gifts there was anN x Nglass chessboard that had a light in each of its cells. When the board was turned on a distinct cell would light up every second, and then go dark.

The cells would light up in the sequence shown in the diagram. Each cell is marked with the second in which it would light up.

(The numbers in the grids stand for the time when the corresponding cell lights up)

In the first second the light at cell (1, 1) would be on. And in the 5th second the cell (3, 1) would be on. Now, Fibsieve is trying to predict which cell will light up at a certain time (given in seconds). Assume thatNis large enough.

Input

Input starts with an integerT (≤ 200), denoting the number of test cases.

Each case will contain an integerS (1 ≤ S ≤ 10¹⁵)which stands for the time.

Output

For each case you have to print the case number and two numbers(x, y), the column and the row number.

Sample Input

Output for Sample Input

Case 1: 2 3

Case 2: 5 4

Case 3: 1 5

计算公式是根据等差数列计算出来的，然后处理一下特殊情况问题就可以了

要把每个走向看成是一个数列值，得到等差数列： 1, 3, 5, 7, 9, ....

然后计算出x和y值来，这里不能使用循环，否则超时。

#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;

void FibsievesFantabulousBirthday()
{
	int T = 0;
	cin>>T;
	long long S = 0;
	for (int i = 0; i < T; i++)
	{
		cin>>S;
		long long rowNum = 1, x = 1, y = 1; 
		y = (long long)sqrt(S);
		S -= (y*y);//代码应用公式：根据等差数列求和变形计算出来
		if (S == 0) S += 1 + 2 *(y-1);//作特殊处理
		else y++;

		if (y % 2 == 0 && S <= y) x = S;
		else if (y % 2 && S <= y)
		{
			x = y;
			y = S;
		}
		else if (y % 2 == 0 && S > y)
		{
			x = y;
			y = y - (S - y );
		}
		else if (y % 2 && S > y)
		{
			x = y - (S - y);
		}
		cout<<"Case "<<i+1<<": "<<x<<' '<<y<<endl;
	}
}
int main()
{
	FibsievesFantabulousBirthday();
	return 0;
}