CF#204DIV2：B. Jeff and Periods

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

• x occurs in sequence a.
• Consider all positions of numbers x in the sequencea (such i, thata_i = 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 integersa₁, a₂, ...,a_n(1 ≤ a_i ≤ 10⁵). The numbers are separated by spaces.

Output

In the first line print integer t — the number of validx. On each of the next t lines print two integers x andp_x, wherex is current suitable value, p_x is the common difference between numbers in the progression (ifx occurs exactly once in the sequence, p_x must equal 0). Print the pairs in the order of increasingx.

Sample test(s)

Input

12

Output

12 0

Input

81 2 1 3 1 2 1 5

Output

41 22 43 05 0

Note

In the first test 2 occurs exactly once in the sequence, ergop₂ = 0.

 

这道题看了我好久才弄懂意思，就是要找出数列中所有下标为等差数列的那些数输出，并输出下标的等差值

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int hash[100005];struct node{    int d;    int pre;    int is;} s[100005];int main(){    int n,i,a,r,cnt,maxn;    while(~scanf("%d",&n))    {        memset(hash,0,sizeof(hash));        memset(s,0,sizeof(s));        maxn = 0;        for(i = 0; i<n; i++)        {            scanf("%d",&a);            maxn = max(maxn,a);//找出最大的那项            if(!hash[a])            {                s[a].pre = i;//将现在的位置                hash[a]++;                s[a].is = 0;            }            else if(hash[a] == 1)            {                r = i - s[a].pre;//计算位置的公差                s[a].d = r;//放入公差                s[a].pre = i;//位置放入                hash[a]++;            }            else if(s[a].is != -1)            {                hash[a]++;                r = i-s[a].pre;                s[a].pre = i;                if(r!=s[a].d)                    s[a].is = -1;//公差不相等，标记为不行            }        }        cnt = 0;        for(i = 1; i<=maxn; i++)        {            if(s[i].is==0 && hash[i])//所有符合的状况                cnt++;        }        printf("%d\n",cnt);        for(i = 1; i<=maxn; i++)        {            if(s[i].is == -1 || !hash[i])                continue;            printf("%d %d\n",i,s[i].d);        }    }    return 0;}