1008 - Fibsieve`s Fantabulous Birthday （lightoj）

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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。

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

1008 - Fibsieve`s Fantabulous Birthday

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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

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);
        }
    }
}