zoukankan      html  css  js  c++  java
  • lightoj--1008--Fibsieve`s Fantabulous Birthday(水题)

    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

    Source

    Problem Setter: Muntasir Muzahid Chowdhury
    Special Thanks: Jane Alam Jan (Dataset)

    #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
    ",i,a);
    			else
    				printf("Case %d: 1 %d
    ",i,a);
    		 } 
    		 else if(b==a+1)
    			 printf("Case %d: %d %d
    ",i,a+1,a+1);
    		else if(b<a+1)
    		{
    			if(a%2==0)
    				printf("Case %d: %d %d
    ",i,a+1,b);
    			else
    				printf("Case %d: %d %d
    ",i,b,a+1);
    		}
    		else 
    		{
    			if(a%2==0)
    				printf("Case %d: %d %d
    ",i,(a+1)*(a+1)-n+1,a+1);
    			else
    				printf("Case %d: %d %d
    ",i,a+1,(a+1)*(a+1)-n+1); 
    		}
    	}
    	return 0;
    }


  • 相关阅读:
    TCP11种状态
    多客户连接僵尸进程的处理
    gethostname(获取主机名)、gethostbyname(由主机名获取IP地址)
    点对点通信实例
    XCTF simple js
    XCTF WEB backup
    bugku SKCTF管理系统
    php漏洞 sha1函数
    bugku--速度要快
    bugku秋名山车神
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273642.html
Copyright © 2011-2022 走看看