ZOJ Problem Set - 1879

ZOJ Problem Set - 1879
Jolly Jumpers

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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.


Input

Each line of input contains an integer n < 3000 followed by n integers representing the sequence.


Output

For each line of input, generate a line of output saying "Jolly" or "Not jolly".


Sample Input

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


Sample Output

Jolly
Not jolly


Source: University of Waterloo Local Contest 2000.09.30

//简单模拟题

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
bool hash[3002];
int a[3002];
int main()
{
    int n,i;
    bool f;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
         scanf("%d",&a[i]),hash[i]=0;
        for(f=1,i=2;i<n;i++)
        {
            if(a[i]>a[i-1])
            {
                if(a[i]-a[i-1]>=n||a[i]-a[i-1]<1)
                   {
                       f=0;break;
                   }
               if(!hash[a[i]-a[i-1]])
                 hash[a[i]-a[i-1]]=1;
                else
                {
                    f=0;break;
                }
            }
            else
            {
               if(a[i-1]-a[i]>=n||a[i-1]-a[i]<1)
                   {
                       f=0;break;
                   }
               if(!hash[a[i-1]-a[i]])
                 hash[a[i-1]-a[i]]=1;
                else
                {
                    f=0;break;
                }
            }
        }
        if(f) printf("Jolly\n");
        else printf("Not jolly\n");
    }
    return 0;
}

posted on 2012-08-01 15:34  江财小子  阅读(152)  评论(0编辑  收藏  举报