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[CF]#579 D3 A. Circle of Students
There are n students standing in a circle in some order. ...
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14
2019/08

[CF]#579 D3 A. Circle of Students

There are n students standing in a circle in some order. The index of the i-th student is pi. It is guaranteed that all indices of students are distinct integers from 1 to n (i. e. they form a permutation).

Students want to start a round dance. A clockwise round dance can be started if the student 2 comes right after the student 1 in clockwise order (there are no students between them), the student 3 comes right after the student 2 in clockwise order, and so on, and the student n comes right after the student n−1 in clockwise order. A counterclockwise round dance is almost the same thing — the only difference is that the student i should be right after the student i−1 in counterclockwise order (this condition should be met for every i from 2 to n).

For example, if the indices of students listed in clockwise order are [2,3,4,5,1], then they can start a clockwise round dance. If the students have indices [3,2,1,4] in clockwise order, then they can start a counterclockwise round dance.

Your task is to determine whether it is possible to start a round dance. Note that the students cannot change their positions before starting the dance; they cannot swap or leave the circle, and no other student can enter the circle.

You have to answer q independent queries.

Input
The first line of the input contains one integer q (1≤q≤200) — the number of queries. Then q queries follow.

The first line of the query contains one integer n (1≤n≤200) — the number of students.

The second line of the query contains a permutation of indices p1,p2,…,pn (1≤pi≤n), where pi is the index of the i-th student (in clockwise order). It is guaranteed that all pi are distinct integers from 1 to n (i. e. they form a permutation).

Output
For each query, print the answer on it. If a round dance can be started with the given order of students, print "YES". Otherwise print "NO".

分析

既然是环 简单一点 2倍数组 一定会组成一个换 首尾相接
当然 也可以 判断 前后数字 abs是否差值为1

AC


import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int q = cin.nextInt();
        for (int i = 0; i < q; i++) {
            int n = cin.nextInt();
            int[] num = new int[2 * n];
            for (int j = 0; j < n; j++) {
                num[j] = cin.nextInt();
            }
            for (int j = n, k = 0; j < n * 2; j++, k++) {
                num[j] = num[k];
            }
//            System.out.println(Arrays.toString(num));
            boolean ans=f(num, n);
            if(ans){
                System.out.println("YES");
            }else
                System.out.println("NO");
        }
    }

    private static boolean f(int[] num, int n) {
        if (n == 1) {
            return true;
        } else {
            int is = 1;
            for (int i = 0; i < 2 * n; i++) {
                if (num[i] == is) {
                    is++;
                    if (is == n + 1) {
                        return true;
                    }
                } else {
                    is = 1;
                }
            }
            is = n;
            for (int i = 0; i < 2 * n; i++) {
                if (num[i] == is) {
                    is--;
                    if (is == 0) {
                        return true;
                    }
                } else {
                    is = n;
                }
            }

        }
        return false;
    }
}
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Last modification:August 14th, 2019 at 12:52 pm
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2 comments

  1. clatterrr

    发现一个用java的大佬OωO

    1. Dxoca
      @clatterrr

      嗯 考虑换c++来写了