1008 - Fibsieve`s Fantabulous Birthday (lightoj)

本文介绍了一种预测特殊棋盘上灯光亮起规律的方法。该棋盘为NxN大小,每个单元格内有一盏灯,按特定序列依次亮起。文章通过分析规律并提供算法实现,帮助读者理解如何根据给定的时间点确定哪一盏灯会亮起。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

http://www.lightoj.com/volume_showproblem.php?problem=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

 找呀找规律

#include <cstdio>
#include <algorithm>
#include <stack>
#include <cmath>
#define SI(T)int T;scanf("%d",&T)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int SIZE=5e4+10;
const int maxn=1<<30;
int main(){
    SI(T);
    for(int cas=1;cas<=T;cas++){
        LL s;
        scanf("%lld",&s);
        LL n=(LL)sqrt(s*1.);
        if(n*n==s){
            if(n&1)printf("Case %d: %d %lld\n",cas,1,n);
            else printf("Case %d: %lld %d\n",cas,n,1);
            continue;
        }
        if(n&1){
            if(n*n+n+1>=s)printf("Case %d: %lld %lld\n",cas,s-n*n,n+1);
            else printf("Case %d: %lld %lld\n",cas,n+1,(n+1)*(n+1)-s+1);
        }
        else {
            if(n*n+n+1>=s)printf("Case %d: %lld %lld\n",cas,1+n,s-n*n);
            else printf("Case %d: %lld %lld\n",cas,(n+1)*(n+1)-s+1,n+1);
        }
    }
}


 


 

资源下载链接为: https://pan.quark.cn/s/f989b9092fc5 今天给大家分享一个关于C#自定义字符串替换方法的实例,希望能对大家有所帮助。具体介绍如下: 之前我遇到了一个算法题,题目要求将一个字符串中的某些片段替换为指定的新字符串片段。例如,对于源字符串“abcdeabcdfbcdefg”,需要将其中的“cde”替换为“12345”,最终得到的结果字符串是“ab12345abcdfb12345fg”,即从“abcdeabcdfbcdefg”变为“ab12345abcdfb12345fg”。 经过分析,我发现不能直接使用C#自带的string.Replace方法来实现这个功能。于是,我决定自定义一个方法来完成这个任务。这个方法的参数包括:原始字符串originalString、需要被替换的字符串片段strToBeReplaced以及用于替换的新字符串片段newString。 在实现过程中,我首先遍历原始字符串,查找需要被替换的字符串片段strToBeReplaced出现的位置。找到后,就将其替换为新字符串片段newString。需要注意的是,在替换过程中,要确保替换操作不会影响后续的查找和替换,避免遗漏或重复替换的情况发生。 以下是实现代码的大概逻辑: 初始化一个空的字符串result,用于存储最终替换后的结果。 使用IndexOf方法在原始字符串中查找strToBeReplaced的位置。 如果找到了,就将originalString中从开头到strToBeReplaced出现位置之前的部分,以及newString拼接到result中,然后将originalString的查找范围更新为strToBeReplaced之后的部分。 如果没有找到,就直接将剩余的originalString拼接到result中。 重复上述步骤,直到originalStr
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值