E-jolly Jumpers

E-jolly Jumpers Time limit:3000 Ms Memory limit:0 KB 64bit Io format:% LLD & % llusubmit status

Description

A sequenceN> 0Integers is calledJolly JumperIf the absolute values of the difference between successive elements take on all the values 1 throughN-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 integerN<= 3000 followedNIntegers 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 35 1 4 2 -1 6

Sample output

JollyNot jolly

Difficult to understand .. (Qaq was thrown into question)

First, enter a positive integer N, followed by N numbers, and the absolute values of each two consecutive digits subtract Form 1 ~ N-1 digits in n-1.

AC code:

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # include <cstdlib> 7 using namespace std; 8 9 const int max = 3050; 10 int A [Max]; 11 int B [Max]; 12 INT N; 13 int main () {14 While (~ Scanf ("% d", & N) {15 for (INT I = 0; I <n; I ++) {16 scanf ("% d ", & A [I]); 17} 18 if (n = 1) {// when n is 1, 19 printf ("Jolly \ n") is output directly; 20 continue; 21} 22 for (INT I = 1; I <n; I ++) {// the absolute values of the two pairs subtract 23 B [I] = ABS (A [I]-A [I-1]); 24} 25 int temp; 26 for (INT I = 1; I <n; I ++) {// bubble sort 27 for (Int J = n-1; j> = I; j --) {28 If (B [J] <B [J-1]) {29 temp = B [J]; 30 B [J] = B [J-1]; 31 B [J-1] = temp; 32} 33} 34} 35 int ans = 0; 36 for (INT I = 1; I <n; I ++) {// number of equal records 37 If (B [I] = I) 38 ans ++; 39} 40 if (ANS = N-1) 41 printf ("Jolly \ n"); 42 else43 printf ("not jolly \ n"); 44} 45 return 0; 46}

 

E-jolly Jumpers