Codeforces Round #333 (Div. 2) B. Approximating a Constant Range (线段树区间最值)

B. Approximating a Constant Range

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., a_n. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |a_i + 1 - a_i| ≤ 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 a_i for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

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, ..., a_n (1 ≤ a_i ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Sample test(s)

input

51 2 3 3 2

output

4

input

115 4 5 5 6 7 8 8 8 7 6

output

5

Note

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].

题意：给你一个序列，找出最长的区间，这个区间的最大值和最小值相差1

思路：i,j两个指针j指向开头，如果一个区间的最大值和最小值相差大于1，那么它所属的区间也一定不满足条件，那么就不必计算，那么j的指针向前进，然后外面的i也不断前进

总结：刚开始觉得线段树写肯定不行，然后开始找规律，并没有找到，就随便写了一个普通的区间线段树交了，TLE了，然后开始在线段树基础上找规律，列了几组数据找规律，最后终于找到。。还是要有耐心。。

ac代码：

#include<stdio.h>#define MAXN 100100int MIN(int a,int b){return a>b?b:a;}int MAX(int a,int b){return a>b?a:b;}struct s{int left;int right;int min;int max;}tree[MAXN*3];int num[MAXN];void build(int l,int r,int i){tree[i].left=l;tree[i].right=r;if(l==r){tree[i].min=num[l];tree[i].max=num[l];}else{int mid;mid=(l+r)/2;build(l,mid,i*2);build(mid+1,r,i*2+1);tree[i].min=MIN(tree[i*2].min,tree[i*2+1].min);tree[i].max=MAX(tree[i*2].max,tree[i*2+1].max);    }}int query1(int i,int l,int r){      if(tree[i].left==l&&tree[i].right==r)      {          return tree[i].min;      }      if(r<=tree[i*2].right)      {          return query1(i*2,l,r);      }      if(l>=tree[i*2+1].left)      {          return query1(i*2+1,l,r);      }      int mid=(tree[i].left+tree[i].right)/2;      return MIN(query1(i*2,l,mid),query1(i*2+1,mid+1,r));  }int query2(int i,int l,int r){      if(tree[i].left==l&&tree[i].right==r)      {          return tree[i].max;      }      if(r<=tree[i*2].right)      {          return query2(i*2,l,r);      }      if(l>=tree[i*2+1].left)      {          return query2(i*2+1,l,r);      }      int mid=(tree[i].left+tree[i].right)/2;      return MAX(query2(i*2,l,mid),query2(i*2+1,mid+1,r));  }int main(){int n,i,j;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&num[i]);build(1,n,1);int ans;int j=1; for(i=1;i<=n;i++){int M=query2(1,j,i);int mi=query1(1,j,i);while(M-mi>1&&j<=i){j++;M=query2(1,j,i);    mi=query1(1,j,i);}ans=MAX(ans,i-j+1);}printf("%d\n",ans);}return 0;}