ZCMU1857: Problem C: Jolly Jumpers

最新推荐文章于 2022-04-04 22:21:10 发布

zjykshen

最新推荐文章于 2022-04-04 22:21:10 发布

阅读量400

点赞数

CC 4.0 BY-SA版权

分类专栏： ZCMU

本文链接：https://blog.youkuaiyun.com/qq_40916491/article/details/78988353

ZCMU 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一个用于判断序列是否为 Jolly Jumpers 的算法实现。通过计算序列中每对相邻元素之间的绝对差值，并确保这些差值包含从1到n-1的所有整数值来完成判断。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Problem C: Jolly Jumpers

A sequence of n > 0 integers is called a jolly jumper if the absolute values of the difference between successive elements take on all the values 1 through n-1. For instance,

1 4 2 3

is a jolly jumper, because the absolutes differences are 3, 2, and 1 respectively. The definition implies that any sequence of a single integer is a jolly jumper. You are to write a program to determine whether or not each of a number of sequences is a jolly jumper.

Each line of input contains an integer n < 3000 followed by n integers representing the sequence. For each line of input, generate a line of output saying "Jolly" or "Not jolly".

Input

Output

Sample Input

4 1 4 2 3
5 1 4 2 -1 6

Sample Output

Jolly
Not jolly

HINT

Source

一开始审题以为只需要满足两个相邻数的差为1||2||3 wa。。。

重新读题 take on all the values 1 through n-1 1到n-1

a放输入 b放差值排序看是不是b[i]=i+1;

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
	int n,a[3500],b[3500];
	while(scanf("%d",&n)!=EOF)
	{
		int i,flag=0;
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		for(i=0;i<n-1;i++)
		   b[i]=abs(a[i]-a[i+1]); 
		sort(b,b+n-1);
		for(i=0;i<n-1;i++)
		{
			if(b[i]!=i+1)
			{
				flag=1;
			    break;
			 } 			
		}
		 if(flag==0)
		 printf("Jolly\n");
		 else
		 printf("Not jolly\n");
	}
	return 0;
}