Light Oj 1008

最新推荐文章于 2022-09-28 20:07:42 发布

转载最新推荐文章于 2022-09-28 20:07:42 发布 · 134 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/codeyuan/p/4254461.html

Fibsieve举办了一场充满惊喜的生日派对，派对上有一块N x N格子的玻璃棋盘，每个格子内都装有光源。当棋盘开启后，灯光会按照特定的序列照亮每个格子。文章探讨了如何预测给定时间内哪个格子会被点亮，涉及数论和数学逻辑。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Fibsieve`s Fantabulous Birthday
Time Limit:500MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit

Status
Description
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
3
8
20
25
Sample Output
Case 1: 2 3
Case 2: 5 4
Case 3: 1 5

/**********************************************

     author   :   Grant Yuan
     time     :   2014.8.5
     algorithm:   数论
     explain  :    对n开平方然后向上取整得到m:
                      如果m为奇数：
                             if（m*m-n<m)    y=m;x=m*m-n+1;
                               else  x=m;y=n-(m-1)*(m-1);

                      如果m为偶数：
                             if(m*m-n<m)    x=m;y=m*m-n+1;
                                else   y=m;x=n-(m-1)*(m-1);

**********************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

long long t,n,x,y;
long long z;
long long m;

int main()
{
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
      scanf("%lld",&n);
      z=sqrt(n*1.0);
      if(z*z==n) m=z;
      else m=z+1;
      if(m%2){
        if(m*m-n<m){
            y=m;x=m*m-n+1;
        }
        else{
            x=m;y=n-(m-1)*(m-1);
        }}
     else{
        if(m*m-n<m){
            x=m;y=m*m-n+1;
        }
        else{
            y=m;x=n-(m-1)*(m-1);
        }
     }
     printf("Case %d: %lld %lld\n",i,x,y);
      }
    return 0;
}

转载于:https://www.cnblogs.com/codeyuan/p/4254461.html