Light oj:1008 - Fibsieve`s Fantabulous Birthday【规律】_light oj - 1008 - fibsieve`s fantabulous birthday -优快云博客

本文探讨了一个关于预测大型玻璃棋盘在特定时间点亮的特定单元问题。通过分析棋盘灯光序列，读者可以了解如何计算并预测在给定秒数时，哪个单元将会被点亮。该文提供了详细的输入输出格式说明，并给出了解题思路和AC代码，适用于对算法和数据结构感兴趣的开发者。

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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

PROBLEM SETTER: MUNTASIR MUZAHID CHOWDHURY

SPECIAL THANKS: JANE ALAM JAN (DATASET)

思路：在里面找规律a{1，4，9，16，25。。。。}它们总是在第ai的算术平方根个小半圈里面，且是在最边缘的位置~~然后细心找规律。。。然后一个个试验纠错就好啦~~~

AC-code:

#include<cstdio>
#include<cmath>
int main()
{
	int t,i,a,b;
	long long n;
	scanf("%d",&t);
	for(i=1;i<=t;i++)
	{
		scanf("%lld",&n);
		a=sqrt(n);
		b=n-a*a;
		if(b==0)
		{
			if(a%2==0)
				printf("Case %d: %d 1\n",i,a);
			else
				printf("Case %d: 1 %d\n",i,a);
		 } 
		 else if(b==a+1)
			 printf("Case %d: %d %d\n",i,a+1,a+1);
		else if(b<a+1)
		{
			if(a%2==0)
				printf("Case %d: %d %d\n",i,a+1,b);
			else
				printf("Case %d: %d %d\n",i,b,a+1);
		}
		else 
		{
			if(a%2==0)
				printf("Case %d: %d %d\n",i,(a+1)*(a+1)-n+1,a+1);
			else
				printf("Case %d: %d %d\n",i,a+1,(a+1)*(a+1)-n+1); 
		}
	}
	return 0;
}