Topic links: https://vjudge.net/problem/284704/origin
Approximating a Constant Range
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.
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, ..., an (1 ≤ ai ≤ 100 000).
Output
Print a single number — the maximum length of an almost constant range of the given sequence.
Examples
5
1 2 3 3 2
4
11
5 4 5 5 6 7 8 8 8 7 6
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].
Title meaning: is to give you an array, the array to ensure that the difference before and after the two elements <= 1, and want you to find any two consecutive numbers differ by no more than a length of the longest string.
Topic Analysis: can have the following practices:
1.dp
2. Optimization of the length of the extended string
3.rmq practice
#include <bits/stdc++.h> using namespace std; const int maxn=1e5+50; int a[maxn],dp[maxn]; int main() { int n; scanf("%d",&n); int ans=0; for( int i=1; i<=n; i++ ) { int x; scanf( "%d",&x ); if( dp[x-1]>dp[x+1] ) ans=max( ans,i-max( dp[x-2],dp[x+1] ) ); else ans=max( ans, i-max(dp[x+2],dp[x-1] ) ); dp[x]=i; } printf( "%d\n",ans ); return 0; }
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e6+10; int n,cnt,ans,a[maxn]; int main() { cin>>n; for(int i=1;i<=n;i++)cin>>a[i]; ANS = 0 ; int L = . 1 , R & lt = 2 ; int One = A [ . 1 ], TWO = A [ 2 ]; // difference before and after the two elements <=. 1 the while (R & lt <= n-) { the while (A [r] == One || A [r] == tWO) r ++; // fixing two elements to expand right l constant r ANS = max (ANS, rl is an); // determine the length of one = a [r- . 1 ]; tWO = A [R & lt]; // update the two elements of L = R- . 1 ; // L update the while (A [L] == One) l - the; // L backtracking L ++ ; } cout<<ans<<endl; return 0; }
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e5+10; int n,cnt,ans,a[maxn],dp[maxn][20],fp[maxn][20]; void init_rmq(){ for(int i=1;i<=n;i++)dp[i][0]=fp[i][0]=a[i]; for(int i=1;(1<<i)<=n;i++){ for(int j=1;j+(1<<i)-1<=n;j++){ dp[j][i]=max(dp[j][i-1],dp[j+(1<<i-1)][i-1]); fp[j][i]=min(fp[j][i-1],fp[j+(1<<i-1)][i-1]); } } } int query_max(int l,int r){ int k=log2(r-l+1); return max(dp[l][k],dp[r-(1<<k)+1][k]); } int query_min(int l,int r){ int k=log2(r-l+1); return min(fp[l][k],fp[r-(1<<k)+1][k]); } int main () { cin>>n; for(int i=1;i<=n;i++)cin>>a[i]; years = 0 ; init_rmq(); int j=1; for(int i=1;i<=n;i++) { while(query_max(j,i)-query_min(j,i)>1&&j<=i) { j++; } years = max (years i-j + 1 ); } cout<<ans<<endl; return 0; }