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;
}
}
本文地址:https://dxoca.cn/Algorithm/253.html 百度已收录
版权说明:若无注明,本文皆为“Dxoca's blog (寒光博客)”原创,转载请保留文章出处。
发现一个用java的大佬OωO
嗯 考虑换c++来写了