Codeforces Round #204 (Div. 2) B - Jeff and Periods

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.

Examples

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

Note

In the first test 2 occurs exactly once in the sequence, ergo p2 = 0.
题意：给你n个数找出同一数的下标和为等差数列的数并输出公差；
思路：开个数组保存数的第一次出现的下标first，在开个数组保存这个数下次出现的下标second，并用个数组记录两者距离
当这个数再次出现时这个下标和上个数和上上个数下标距离是否相同，不同标记这个数为no，注意更新first和second，还有这个数只出现一次时也是等差数列。

代码如下：

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
int a[100005],first[100005],second[100005],sum[100005],flag[100005];
int main()
{
int n;
while(~scanf("%d",&n)){
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
memset(first,-1,sizeof(first));
memset(second,-1,sizeof(second));
memset(sum,0,sizeof(sum));
memset(flag,0,sizeof(flag));
int cnt=0;
for(int i=0;i<n;i++)
{
if(flag[a[i]]) continue;
if(first[a[i]]==-1)
{
first[a[i]]=i;
}
else if(second[a[i]]==-1)
{
second[a[i]]=i;
sum[a[i]]=(i-first[a[i]]);
}
else
{
int ans=(i-second[a[i]]);
if(ans!=sum[a[i]])
{
flag[a[i]]=1;cnt++;
}
else
{
first[a[i]]=second[a[i]];
second[a[i]]=i;
}
}
}
sort(a,a+n);
int pos=0;
for(int i=0;i<n;i++)
{
if(a[i]!=a[i-1]) pos++;
}
printf("%d\n",pos-cnt);
for(int i=0;i<n;i++)
{
if(a[i]!=a[i-1])
{
if(flag[a[i]]==0)
{
printf("%d %d\n",a[i],sum[a[i]]);
}
}
}
}
}