zoukankan      html  css  js  c++  java
  • Light Oj 1008

    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
    ",i,x,y);
          }
        return 0;
    }
    


     

  • 相关阅读:
    欢乐送小程序自动化探索实践
    看完这篇还不了解 Nginx,那我就哭了!
    测试人的技术栈
    Bug,项目过程中的重要数据
    什么是测试开发工程师?
    hdu 1219 AC Me
    hdu 1202 The calculation of GPA(算绩点问题)
    hdu1205吃糖果(插空法)
    hdu1201(18岁生日)
    hdu1231最大连续子序列
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254461.html
Copyright © 2011-2022 走看看