light oj 1008 Fibsieve`s Fantabulous Birthday （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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本文介绍了一个有趣的编程问题：根据特定的数字序列填充棋盘，并找出指定时间点亮的单元格位置。通过观察序列规律，利用数学方法计算出目标单元格的行列坐标。

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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

Sample Output

Case 1: 2 3

Case 2: 5 4

Case 3: 1 5

题意：按照上述图片里数字的填充方式，给你一个数字，求它所在的列和行。

解题思路：规律题。填充的方式，可看做一个个正方形边框，以左下角为中心。（例如，第一个正方形边框的数字有1，第二个有1 2 3 4，第三个有1 2 9 8 7 6 5 4）.每给出一个数字，可先算出这个数字在哪个正方形边框。然后再计算行和列，分奇偶来计算就可以了，注意计算过程中防止数据类型溢出，要用long long。

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
//ifstream in("e://in.txt");
//ofstream out("e://out.txt");

int main()
{
    ll s,i,j;
    int t;
    cin>>t;
    for(int cas=1; cas<=t; cas++)
    {
        cin>>s;
        ll num=sqrt(s);
        if(num*num!=s)
            num++;
        ll flag=num*num-(num-1);
        if(num%2!=0)
        {
            if(s>flag)
                i=abs(num*num-s)+1,j=num;
            else
                i=num,j=abs((num-1)*(num-1)+1-s)+1;
        }
        else
        {
            if(s>flag)
                i=num,j=abs(num*num-s)+1;
            else
                i=abs((num-1)*(num-1)+1-s)+1,j=num;
        }
        cout<<"Case "<<cas<<": "<<i<<' '<<j<<endl;
    }
    return 0;
}