UVa OJ - 10038 Jolly Jumpers

The following program is my ACcepted code for UVA-10038 .
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know.  :D

此乃UVA 10038 的AC code!

歡迎一同討論學習，如有錯誤與任何建議請留言 : )

點這裡看題目 Click here to see this Problem!

//This program is for UVA 10038 Jolly Jumpers
//題目來源 Problem link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=12

//還沒AC!!!!!

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
    int n, seq[3001];
    bool visit[3001];
    bool ans;
 
    while(scanf("%d", &n) != EOF)
    {
        memset(visit, false, sizeof(visit));
     
        for(int i = 0; i < n; i++)
            scanf("%d", &seq[i]);
     
     
        for(int i = 0; i < n-1; i++)
        {
            int tmp = abs(seq[i] - seq[i + 1]);
            visit[tmp] = true;
        }
     
        ans = true;
        for(int i = 1; i < n; i++)
            if(!visit[i])//was not visited
                ans = false;
             
        if(ans)
            printf("Jolly\n");
        else
            printf("Not jolly\n");
    }
    return 0;
}

Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D 

歡迎使用與分享任何內容，但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )