CodeForces 352B 水题

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，查找一串序列中，出现位置为等差数列的数

代码：//By Sean Chen
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;

struct node{
int num,p;
};
map <int,int> P;
map <int,int> prepos;
map <int,int> flag;
vector <int> num;
node ans[100005];
int b;
int cmp(node a,node b)
{
return a.num<b.num;
}
int main()
{
int n;
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&b);
if (!prepos[b])
{
num.push_back(b);
flag[b]=1;
}
else if (P[b]!=0 && i-prepos[b]!=P[b])
{
flag[b]=0;
}
else if (P[b]==0)
{
P[b]=i-prepos[b];
}
prepos[b]=i;
}
node temp;
int cnt=0;
for (int i=0;i<=num.size();i++)
{
if (flag[num[i]]==1)
{
temp.num=num[i];
temp.p=P[num[i]];
ans[cnt++]=temp;
}
}
sort(ans,ans+cnt,cmp);
printf("%d\n",cnt);
for (int i=0;i<cnt;i++)
{
printf("%d %d\n",ans[i].num,ans[i].p);
}
return 0;
}