When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.
A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.
Find the length of the longest almost constant range.
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).
Print a single number — the maximum length of an almost constant range of the given sequence.
5 1 2 3 3 2
4
11 5 4 5 5 6 7 8 8 8 7 6
5
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
题目大意：
给你一串数，让你求最大的区间（最大值-最小值<=1）
解题思路：
我是用集合做的。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5+5;
const int mod = 1e9+7;
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
LL gcd(LL a, LL b)
{
if(b == 0)
return a;
return gcd(b, a%b);
}
multiset <int> s;
int arr[maxn];
int main()
{
int n;
while(cin>>n)
{
s.clear();
for(int i=0; i<n; i++)
scanf("%d",&arr[i]);
int i , j, ret = -INF;
i = j = 0;
while(j < n)
{
s.insert(arr[j]);
if(abs(*s.begin()-*s.rbegin()) > 1)
{
s.erase(s.find(arr[i]));
i++;
}
ret = max(ret, int(s.size()));
j++;
}
cout<<ret<<endl;
}
return 0;
}
