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.

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

Note

In the first test 2 occurs exactly once in the sequence, ergo p2 = 0.

开始把题读错了，题目意思是如果不成等差数列就去掉！

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
#define M 105000
int pri[M],p[M],answer[M];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF){
memset(p,-1,sizeof(p));
memset(answer,-1,sizeof(answer));
for(i=0;i<n;i++)
scanf("%d",&pri[i]);
for(i=0;i<n;i++){
if(answer[pri[i]]==-2)
continue;
if(p[pri[i]]==-1){
p[pri[i]]=i;
}
else if(p[pri[i]]!=-1&&answer[pri[i]]==-1){

answer[pri[i]]=i-p[pri[i]]; p[pri[i]]=i;
}
else if(p[pri[i]]!=-1&&answer[pri[i]]!=-1){
if(i-p[pri[i]]!=answer[pri[i]]){
answer[pri[i]]=-2;
}
else
p[pri[i]]=i;
}
}
int k=0;
for(i=0;i<M;i++){
if(p[i]!=-1&&answer[i]!=-2)
k++;
}
printf("%d\n",k);
for(i=0;i<M;i++){
if(p[i]!=-1&&answer[i]!=-2&&answer[i]!=-1)
printf("%d %d\n",i,answer[i]);
else if(answer[i]==-1&&p[i]!=-1)
printf("%d 0\n",i);
}
}
return 0;
}