B. Jeff and Periods(cf)

B. Jeff and Periods
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

One day Jeff got hold of an integer sequence a1, a2, ..., an of length n. The boy immediately decided to analyze the sequence. For that, he needs to find all values of x, for which these conditions hold:

  • x occurs in sequence a.
  • Consider all positions of numbers x in the sequence a (such i, that ai = x). These numbers, sorted in the increasing order, must form an arithmetic progression.

 

Help Jeff, find all x that meet the problem conditions.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains integers a1, a2, ..., an (1 ≤ ai ≤ 105). The numbers are separated by spaces.

Output

In the first line print integer t — the number of valid x. On each of the next t lines print two integers x and px, where x is current suitable value, px is the common difference between numbers in the progression (if x occurs exactly once in the sequence, px must equal 0). Print the pairs in the order of increasing x.

Sample test(s)
Input
1
2
Output
1
2 0
Input
8
1 2 1 3 1 2 1 5
Output
4
1 2
2 4
3 0
5 0
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <vector>
 4 #include <algorithm>
 5 const int Max=100005;
 6 using namespace std;
 7 int main()
 8 {
 9     int n,x;
10     int vis[Max],p[Max];
11     while(~scanf("%d",&n))
12     {
13         int k = 0,i,j;
14         vector<int>G[Max];
15         memset(vis,0,sizeof(vis));
16         for (i = 0; i < n; i++)
17         {
18             scanf("%d",&x);
19             G[x].push_back(i);//将所有x的位置存入vector中
20             if (!vis[x])
21             {
22                 vis[x] = 1;
23                 p[k++] = x;
24             }
25 
26         }
27         int cnt = 0;
28         memset(vis,-1,sizeof(vis));
29         for (j = 0; j < k; j++)
30         {
31             int len = G[p[j]].size();
32             if (len==1)
33             {
34                 ++cnt;
35                 vis[p[j]]= 0;
36                 continue;
37             }
38             int d = G[p[j]][1]-G[p[j]][0];//求公差
39             for (i = 1; i < len; i++)
40             {
41                 if (G[p[j]][i]-G[p[j]][i-1]!=d)
42                     break;
43             }
44             if (i >= len)//说明p[j]的各位置是等差数列
45             {
46                 ++cnt;
47                 vis[p[j]] = d;//表示p[j]各位置的公差为d
48             }
49         }
50         printf("%d\n",cnt);
51         sort(p, p+k);
52         for (i = 0; i < k ; i++)
53         {
54             if (vis[p[i]]!=-1)
55             {
56                 printf("%d %d\n",p[i],vis[p[i]]);
57             }
58         }
59     }
60     return 0;
61 }
View Code

 

posted @ 2013-10-05 16:35  N_ll  阅读(288)  评论(0)  编辑  收藏  举报