zoukankan      html  css  js  c++  java
  • light oj 1008

    1008 - Fibsieve`s Fantabulous Birthday
    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

    3

    8

    20

    25

    Case 1: 2 3

    Case 2: 5 4

    Case 3: 1 5

     题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1008

    算是道模拟题吧!找出数的排列顺序模拟下即可

    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #include<stdlib.h>
    #define LL long long 
    #define DD double
    #define MAX 20000000
    using namespace std;
    int main()
    {
    	int t,k;
    	DD n;
    	LL m,j,i;
    	scanf("%d",&t);
    	k=1;
    	while(t--)
    	{
    		scanf("%lf",&n);
    		DD ans=sqrt(n);
    		LL ant=(LL)(ans);
    		printf("Case %d: ",k++);
    		if(ans==ant)
    		{
    			if(ant%2==0)
    			printf("%lld 1
    ",ant);
    			else if(ant&1)
    			printf("1 %lld
    ",ant);
    		}
    		else
    		{
    			LL sum=pow(ant+1,2);
    			LL sun=pow(ant,2);
    			LL flag=(LL)(sum-n);
    			if(flag==ant)
    			    printf("%lld %lld
    ",ant+1,ant+1);
    			else if(flag<ant)
    			{
    				if((ant+1)%2==0)
    					printf("%lld %lld
    ",ant+1,flag+1);
    				else
    				    printf("%lld %lld
    ",flag+1,ant+1);
    			}
    			else
    			{
    				if((ant+1)%2==0)
    				    printf("%lld %lld
    ",(sum-sun)-flag,ant+1);
                    else
                        printf("%lld %lld
    ",ant+1,(sum-sun)-flag);
    			}
    		}
    	}
    	return 0;
    } 
    

      

  • 相关阅读:
    @Autowired mapper 层次 bean 带红线
    java 类加载机制 阿里面试题
    liunx 修改ssh 端口22
    通过mysqlbinlog 恢复数据
    网页命名规则
    子选择符 、相邻选择符 、 兄弟选择符 、 伪类选择符
    css的一些基础知识
    HTML5的表单所有type类型
    一行代码解决各种IE兼容问题,IE6,IE7,IE8,IE9,IE10(复制)
    网站开发最常用的代码(复制)
  • 原文地址:https://www.cnblogs.com/tonghao/p/5016191.html
Copyright © 2011-2022 走看看