light OJ1008 - Fibsieve`s Fantabulous Birthday 【找规律】

最新推荐文章于 2018-05-30 15:28:33 发布

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最新推荐文章于 2018-05-30 15:28:33 发布

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本文链接：https://blog.youkuaiyun.com/liu6886/article/details/50153531

本文介绍了一种特殊的NxN玻璃棋盘，每个格子内有一盏灯，这些灯会按照特定的时间序列亮起和熄灭。文章提供了一个算法，用于预测在任意给定秒数时，哪一格的灯会被点亮。

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1008 - Fibsieve`s Fantabulous Birthday

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

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 an N x N glass 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 that N is large enough.

Input

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

Each case will contain an integer S (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

在C语言的库函数中，floor函数的语法如下：
#include <math.h>
double floor( double arg );
功能：函数返回参数不大于arg的最大整数。例如,
x = 6.04;
y = floor(x)=6.0;
y的值为6.0
与floor函数对应的是ceil函数，即上取整函数。
ceil函数
ceil函数的作用是求不小于给定实数的最小整数。
ceil(2)=ceil(1.2)=ceil(1.5)=2.00
使用该函数需要包含头文件<math.h>,该函数返回值为double型。

#include<cstdio>
#include<cmath>
#include<cstring>
int main(){
	int T;
	scanf("%d",&T);
	for(int t = 1; t <= T; t++){
		long long  x, y;
		long long  n, s;
		scanf("%lld", &s);
		n = ceil(sqrt(s*1.0));
		if(n*n - s < n){
				x = n;
				y = n*n-s+1;
			}
		else{
			y = n;
			x = s-(n-1)*(n-1);
		}
		if(n&1){
			long long  d = x;
			x = y;
			y = d;
		}
		printf("Case %d: %d %d\n", t, x, y);
	}
	return 0;
}