题目描述
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.
Each line of input contains an integer n < 3000 followed by n integers representing the sequence. For each line of input, generate a line of output saying "Jolly" or "Not jolly".
样例输入
4 1 4 2 3
5 1 4 2 -1 6
样例输出
Jolly
Not jolly
解析：题目就是问输入的N个数，两两相邻的差值，最后是否1到N-1每个数都有。所以我们可以直接利用数组来解决。
#include <stdio.h>
#include <math.h>
int k[3000];//存储输入的数
int h[10005];//存储差值，差值为2就存入k【2】,如此
int main()
{int n,i,shi;while(~scanf("%d",&n)){for(i=0;i<10005;i++) h[i]=0; //清空一下 for(i=0;i<n;i++) scanf("%d",&k[i]);for(i=0;i<n-1;i++) h[abs(k[i]-k[i+1])]++; //注意绝对值 shi=1;//shi=1代表该值存在 for(i=1;i<=n-1;i++){if(h[i]==0){ //等于0就是说不存在当前i的数，就不满足题意 shi=0;break;}}if(shi==0) printf("Not jolly\n");if(shi==1) printf("Jolly\n");}return 0;
}