【解題】Jolly Jumpers

ACM Volume C 10038 - Jolly Jumpers


The Problem

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.


The Input

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


The 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


解題思考

　　在這一題，若是輸入大小為 n，我們需要先初始化一個大小同樣為 n 的布林(boolean)陣列為非(false)。

　　接著，在輸入每一個數字之後，我們需要將相鄰兩數相減。若是兩數相減之絕對值落在 1 到 n - 1，就將相對應的布林陣列元素設為真(true)。

　　此外，因為題目要求的是：相鄰兩數其差的絕對值恰好為 1 到 n - 1，所以若是其對應的布林陣列元素早已為真，就代表其差的絕對值有重複，也就不能達到題目所求的「恰好為 1 到 n - 1」。換句話說，碰上這種情況，就代表 Not jolly 了。為了快速起見，就可以不用再做接下來的判斷。


參考解答(C++)

#include <iostream>

using namespace std;

int main(void)
{
    int n;
    while (cin >> n)
    {
        int *num = new int[n];
        bool *seen = new bool[n];
        bool jolly = true;

        for (int i = 0; i < n; i++)
        {
            seen[i] = false;
        }

        for (int i = 0; i < n; i++)
        {
            cin >> num[i];

            if (i && jolly)
            {
                int r = num[i] - num[i - 1];

                if (r < 0) { r *= -1; }

                if (r > n - 1 || !r || seen[r])
                {
                    jolly = false;
                }
                else { seen[r] = true; }
            }
        }

        if (jolly)  { cout << "Jolly" << endl; }
        else        { cout << "Not jolly" << endl; }

        delete [] num;
        delete [] seen;
    }

#ifndef ONLINE_JUDGE
    system("pause");
#endif
}