POJ 2575 Jolly Jumpers（我的水题之路——数组绝对差值为1到n-1）

Jolly Jumpers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13365		Accepted: 4128

Description

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

Waterloo local 2000.09.30

对于输入的n个数字，它们的相邻两个元素之间取绝对差值，如果这些差值为1到n-1（无论次序），就输出“Jolly”，否则输出“Not jolly”。

用一个数组subs作为标记数组，标记已经出现过差值，当出现一个差值就标记为1。当所有的差值计算完毕之后，搜索1到n-1的差值是否全部出现，如果是， 输出“Jolly”，否则输出“Not jolly”。优化：因为是n个数字，有n-1个差值，如果这些差值要取满1到n-1，则说明这些差值之间不会有重复的数字，所以在计算差值的过程中就可以进行判断，如果有重复的差值，则立刻判定“Not jolly”。

注意点：
1）计算差值的时候，要取绝对值，可以使用<cmath>里的abs();

代码（1AC）：

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

int subs[3100];
int num[3100];

int main(void){
    int n, i, j;
    int flag;

    while (scanf("%d", &n) != EOF){
        memset(subs, 0, sizeof(subs));
        flag = 1;
        for (i = 0; i < n; i++){
            scanf("%d", &num[i]);
            if (i){
                if (!subs[abs(num[i] - num[i - 1])]){
                    subs[abs(num[i] - num[i - 1])] = 1;
                }
                else{
                    flag = 0;
                }
            }
        }
        for (i = 1; i < n && flag; i++){
            if (subs[i] != 1){
                flag = 0;
            }
        }
        if (flag){
            printf("Jolly\n");
        }
        else{
            printf("Not jolly\n");
        }
    }
    return 0;
}