codeforces-Jeff and Periods

Jeff and Periods

Description

One day Jeff got hold of an integer sequence a₁, a₂, ..., a_n 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 a_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 integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 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 p_x, where x is current suitable value, p_x is the common difference between numbers in the progression (if x occurs exactly once in the sequence, p_x must equal 0). Print the pairs in the order of increasing x.

Sample Input

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

 

题目大意：长度为n的序列a，x在序列a中，a中所有值为x的元素，如果下标构成等差数列，就输出这个元素的值和公差，否则不输出此元素。如果这个元素只出现了一次，则输出这个元素的值和0。在output的第一行输出一共有多少个下标能构成等差数列的元素。

 

这个题目一开始没读懂，后来才明白是找值相同的数，它们的下标构成的等差数列的公差。

（For that, he needs to find all values of x, Consider all positions of numbers x in the sequence a (such i, that a_i = x). These numbers, sorted in the increasing order,（原来这的递增顺序是指下标在递增） must form an arithmetic progression.）

 

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

#define maxn 100005

int step[maxn]; 
int first[maxn];
int last[maxn];

int main()
{
    int i, n, a, c = 0;
    memset(first, -1, sizeof(first));
    memset(last, -1, sizeof(last));
    memset(step, 0, sizeof(step));
    while (scanf("%d", &n) != EOF)
    {
        for (i = 0; i < n; i++)
        {
            scanf("%d", &a);
            if (first[a] == -1)               //取a的值为数组下标，a为元素的值
            {
                first[a] = i;                //first[a]为【第一次】出现 之前未出现的元素 的位置
                c++;                        //只要是第一次出现此数，就计数
            }
            else 
            {
                if (step[a] == 0)
                {
                    step[a] = i - first[a];       //【第一次】出现 某个元素 据第一个与它值相同的元素的步数（即公差）
                }
                else if (step[a] > 0 && (i - last[a]) != step[a])        //如果之后出现的元素 据 上一个与它相同的元素的步数 != 公差
                {
                    step[a] = -1;                                    //则置为-1，不输出此元素
                    c--;                                            //不计入此数
                }
            }
            last[a] = i;                                   //记录这次该元素的位置，以便下次此值再出现时比较步数与公差
        }
        printf("%d
", c);
        for (i = 1; i <= maxn - 1; i++)
        {
            if (first[i] >= 0)
            {
                if (step[i] >= 0)                          //对于只出现一次的数，step[i] = 0，也会输出
                {
                    printf("%d %d
", i, step[i]);          //i为元素a的值，step[i]为公差
                }
            }
        }
    }
    return 0;
}