stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }



const ll INF=(1LL<<60);



int main(void){

    int n; cin >> n;

    ll h[100100];

    ll dp[100100];

    for(int i=1; i<=n; i++) cin >> h[i];



    for(int i=0; i<100100; i++) dp[i]=INF;

    dp[1]=0;



    for(int i=1; i<=n; i++){

        chmin(dp[i+1], dp[i]+abs(h[i]-h[i+1]));

        chmin(dp[i+2], dp[i]+abs(h[i]-h[i+2]));

    }

    cout << dp[n] << endl;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define repr(i,a,b) for(int i=a;i<b;i++)

#define rep(i,n) for(int i=0;i<n;i++)

#define reprrev(i,a,b) for(int i=b-1;i>=a;i--)

#define reprev(i,n) reprrev(i,0,n)

typedef long long ll;

typedef unsigned long long ull;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

 

/* attention

    long longのシフト演算には気をつけよう

    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも

    sizeは(int)とキャストしよう

    ごちゃごちゃ場合分けを考える前に全探索は考えましたか？

    詰まった時に、別の分野の問題として考え直す（これdpでは？、グラフとしてみればいいのでは？）

    多くの問題はパターンマッチだが、パターンに落とし込むまでが難しく、そのための訓練としてわからない問題をあれこれ色々な角度から考察してみるのではないか

    cin.tie(0);

    ios::sync_with_stdio(false);<- これら、printfとかと併用しない方が良さそう



*/



const ll mod = 1e9+7;



void chmod(ll &M){

    if(M >= mod) M %= mod;

    else if(M < 0){

        M += (abs(M)/mod + 1)*mod;

        M %= mod;

    }

}



int getl(int i, int N) { return i==0? N-1:i-1; };

int getr(int i, int N) { return i==N-1? 0:i+1; };



/* <--------------------------------------------> */





int main(void){

    cin.tie(0);

    ios::sync_with_stdio(false);



    int n; cin >> n;

    vector<int> h(n+5, 0);

    repr(i, 1, n+1) cin >> h[i];



    vector<int> dp(n+5, 0);

    repr(i, 2, n+1){

        if(i==2) dp[i]=dp[i-1]+abs(h[i]-h[i-1]);

        else dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]));

    }



    cout << dp[n] << endl;



    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



const int INF = 1<<30;



int dp[100010];

int h[100010];



int main(){

	int n;	cin>>n;

	for(int i=0;i<n;i++) cin>>h[i];



	for(int i=0;i<100010;i++) dp[i]=INF;

	dp[0]=0;





	for(int i=0;i<n;i++){

		dp[i+1] = min(dp[i+1], dp[i]+abs(h[i+1]-h[i]));

		dp[i+2] = min(dp[i+2], dp[i]+abs(h[i+2]-h[i]));

	}



	cout<<dp[n-1]<<endl;





	return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;



#define INF (1<<30)

#define MAX 100010



int dp[MAX];

int h[MAX];



int main(){

	int n;	cin>>n;

	for(int i=0;i<n;i++){

		cin>>h[i];

		dp[i]=INF;

	}

	dp[0]=0;



	for(int i=0;i<n;i++){

		dp[i+1]=min(dp[i+1], dp[i]+abs(h[i+1]-h[i]));

		dp[i+2]=min(dp[i+2], dp[i]+abs(h[i+2]-h[i]));

	}



	cout<<dp[n-1]<<endl;

	



	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int min(int a, int b) {

  if (a < b) return a;

  else return b;

}

int ze(int a) {

  int t = min(a, 0);

  int j = a - 2 * t;

  return j;

}

int main () {

  int N;

  cin >> N;

  int h[1100000];

  for (int i = 0; i < N; i ++) {

    cin >> h[i];

  }

  int cost[100000];

  cost[2] = ze(h[1] - h[0]);

  int k = cost[2] + ze(h[2] - h[1]);

  int j = ze(h[2] - h[0]);

  cost[3] = min(k, j);

  for(int i = 4; i <= N; i ++) {

    int a = cost[i - 1] + ze(h[i - 1] - h[i - 2]);

    int b = cost[i - 2] + ze(h[i - 1] - h[i - 3]);

    cost[i] = min(a, b);

  }

  cout << cost[N] << endl;

}

    


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

int ze(int a, int b) {

  return max(a - b, b - a);

}

int main () {

  int N;

  cin >> N;

  vector<int>he(N);

  for (int i = 0; i < N; i ++) cin >> he[i];

  vector<int>dp(N);

  dp[0] = 0;

  dp[1] = ze(he[0], he[1]);

  for (int i = 2; i < N; i ++) {

    int ku = dp[i - 1] + ze(he[i], he[i - 1]);

    int ma = dp[i - 2] + ze(he[i], he[i - 2]);

    dp[i] = min(ku, ma);

  }

  cout << dp[N - 1] << endl;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



typedef long long int ll;

typedef long double ld;



#define Pb push_back

#define Mp make_pair



#define ff first

#define ss second



#define Max(x,y,z) max(x,max(y,z))

#define Min(x,y,z) min(x,min(y,z))



#define INF LONG_LONG_MAX

#define MINF LONG_LONG_MIN



#define endl "\n"

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)



#define db1(x) cout<<#x<<" : "<<x<<endl;

#define db2(x,y) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<endl;

#define db3(x,y,z) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<"\t"<<#z<<" : "<<z<<endl;



const int N = 1e5+5;

int h[N];

int n;

int dp[N];



int solution(int idx)

{

  if(idx == n)

  {

    return 0;

  }

  if(idx>n)

  {

    return 10000000;

  }

  if(dp[idx] != -1)

  {

    return dp[idx];

  }

  int &ans = dp[idx];

  int dp1=INT_MAX;

  int dp2=INT_MAX;

  if(idx+1 <= n) dp1 = solution(idx+1)+abs(h[idx]-h[idx+1]);

  if(idx+2 <= n) dp2 = solution(idx+2)+abs(h[idx]-h[idx+2]);

  ans = min(dp1,dp2);

  return ans;

}



void solve()

{

    cin >> n;

    for(int i=1;i<=n;i++) cin >> h[i];

    memset(dp,-1,sizeof(dp));

    int ans = solution(1);

    cout << ans;

}



signed main()

{

  IOS;

  //freopen("input.txt", "rt", stdin);

  //freopen("output.txt", "wt", stdout);

  int t=1;

  //cin >> t;

  while(t--)

  {

    solve();

  }

  return 0;

}

/*==============================================================================================================================================================*/


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;



typedef long long int ll;

typedef long double ld;



#define Pb push_back

#define Mp make_pair



#define ff first

#define ss second



#define Max(x,y,z) max(x,max(y,z))

#define Min(x,y,z) min(x,min(y,z))



#define INF LONG_LONG_MAX

#define MINF LONG_LONG_MIN



#define endl "\n"

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)



#define db1(x) cout<<#x<<" : "<<x<<endl;

#define db2(x,y) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<endl;

#define db3(x,y,z) cout<<#x<<" : "<<x<<"\t"<<#y<<" : "<<y<<"\t"<<#z<<" : "<<z<<endl;



void solve()

{

    int n;

    cin >> n;

    int h[n];

    for(int i=0;i<n;i++)  cin >> h[i];

    int dp[n];

    dp[0] = 0;

    dp[1] = abs(h[1]-h[0]);

    for(int i=2;i<n;i++)

    {

      dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]) , dp[i-2]+abs(h[i]-h[i-2]));

    }

    cout << dp[n-1];

}



signed main()

{

  IOS;

  //freopen("input.txt", "rt", stdin);

  //freopen("output.txt", "wt", stdout);

  int t=1;

  //cin >> t;

  while(t--)

  {

    solve();

  }

  return 0;

}

/*==============================================================================================================================================================*/


// Fastest known solution for this problem:
// Original C++ solution (slower):
//**KEEP IT SHORT AND SIMPLE**



#include <bits/stdc++.h>

#define IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define FILES freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

#define REP(i,a,b) for (int i = a; i < b; i++)

#define NL '\n'

#define F first

#define S second

using namespace std;

typedef long long ll;

typedef bool bl;

typedef vector < vector<long long> > vvl;

typedef vector <long long> vll;

typedef pair <long long, long long> pll;

typedef set <long long> sll;

typedef map <long long, long long> mll;

typedef map <long long, bool> mlb;

typedef priority_queue<long long, vector<long long>, greater<long long> > minPq;

double pi = 2 * acos(0.0);

int OO = 0x3f3f3f3f;

long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }

long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }

const int NMAX = 100010;;

ll n, arr[NMAX], dp[NMAX];

ll solve(ll i)

{

    if(i == n - 1)  return 0;

    if(~dp[i])      return dp[i];

    return dp[i] = min(solve(i + 1) + abs(arr[i + 1] - arr[i]), i + 2 <= n - 1 ? solve(i + 2) + abs(arr[i + 2] - arr[i]) : OO);

}

int main()

{

    IO

    cin >> n;

    REP(i, 0, n)    cin >> arr[i];

    memset(dp, -1, sizeof dp);

    cout << solve(0);

    return 0;

}


// Optimized solution to the same problem (faster):
//**KEEP IT SHORT AND SIMPLE**



#include <bits/stdc++.h>

#define IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define FILES freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

#define REP(i,a,b) for (int i = a; i < b; i++)

#define NL '\n'

#define F first

#define S second

#define dot(a, b) (conj(a) * (b)).real()

#define cross(a, b) (conj(a)* (b)).imag()

using namespace std;

typedef long long ll;

typedef long double ld;

typedef bool bl;

typedef vector < vector<long long> > vvl;

typedef vector <long long> vll;

typedef pair <long long, long long> pll;

typedef map <long long, long long> mll;

typedef map <long long, bool> mlb;

typedef priority_queue<long long, vector<long long>, greater<long long> > minPq;

typedef tuple<ll, ll, ll> line;

typedef complex<double> point;

const double pi = 2 * acos(0.0);

const int OO = 0x3f3f3f3f;

int main()

{

    IO

    ll n;

    cin >> n;

    ll h[n], ans[n];

    REP(i, 0, n)    cin >> h[i];

    ans[0] = 0;

    ans[1] = abs(h[1] - h[0]);

    REP(i, 2, n)    ans[i] = min(abs(h[i] - h[i - 1]) + ans[i - 1], abs(h[i - 2] - h[i]) + ans[i - 2]);

    cout << ans[n - 1];

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF+5];

long long mem[INF];



long long rec(int pos)

{

    if(pos==n-1)

    {

        return 0;

    }

    if(mem[pos]!=-1) return mem[pos];

    long long tot=maxx;

    long long left=abs(a[pos]-a[pos+1]);

    long long right=abs(a[pos]-a[pos+2]);

    if(pos+1<n)

    {

        tot=min(tot,left+rec(pos+1));

    }

    if(pos+2<n)

    {

        tot=min(tot,right+rec(pos+2));

    }

    return mem[pos]=tot;

}



int main()

{

    memset(a,0,sizeof(a));

    memset(mem,-1,sizeof(mem));

    scanf("%d",&n);

    for(int i=0;i<n;i++)

    {

        scanf("%lld",&a[i]);

    }

    long long ans=rec(0);

    printf("%lld\n",ans);

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

#define inN(arr ,  n) for(int i = 0 ; i< n ; ++i)  cin>>arr[i];

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

#define all(v)                   ((v).begin(),(v).end())

#define sz(v)                  (int)((v).size())

#define clr(v,d)                memset(v , d , sizeof(v))

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

    for (int j = 1; j <n ; ++j) {

        if(j>=2){

            dp[j] = min(dp[j-1] + abs(h[j-1] - h[j]) , abs(h[j-2] - h[j])+dp[j-2]);

        }

        else if(j ==1){

            dp[j] = abs(h[j] - h[j-1])+dp[j-1];

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

//    cout << solve(0, 1)<<endl;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],dp[MAX];

int solve(int idx)

{

    if(idx==N)return 0;

    if(idx>N)return 1e9;

    if(dp[idx]!=-1)return dp[idx];

    return dp[idx]=min(abs(A[idx+1]-A[idx])+solve(idx+1),abs(A[idx+2]-A[idx])+solve(idx+2));

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

    for(int i=1;i<=N;i++){

        cin >> A[i];

    }

    cout<<solve(1)<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound



typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;



const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;



int dp[100005],h[100005],n,k=2;



signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    memset(dp,0x3f3f,sizeof dp);

    dp[1]=0;

    h[0]=0;

    cin>>n;

    for(int i=1;i<=n;++i) cin>>h[i];

    for(int i=2;i<=n;++i) {

        for(int j=1;j<=min(k,i);++j) {

            dp[i]=min(dp[i],dp[i-j]+abs(h[i]-h[i-j]));

        }

    }

    cout<<dp[n];

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

#include<cmath>

#include<stack>

#include<cstring>

#include<algorithm>

using namespace std;

int n;

int h[100005];

int dp[100005];



int main()

{

memset(dp, 0x3f ,sizeof dp);

scanf("%d",&n);

for( int i = 1 ; i <= n ; i++) scanf("%d",&h[i]);

dp[1] = 0;

for( int i = 1 ; i <= n ; i++)

{

	dp[i+1] = min(dp[i+1],dp[i]+abs(h[i] - h[i+1]));

	dp[i+2] = min(dp[i+2],dp[i]+abs(h[i] - h[i+2]));

}

printf("%d",dp[n]);



	return 0;

}






// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<vector>

#include<set>

#include<list>

#define ll long long

using namespace std;

ll h[200050],dp[200050];

int main()

{

	ll n;

	cin>>n;

	for(int i=1;i<=n;i++)

		cin>>h[i];

	memset(dp,0,sizeof(dp));

	dp[2]=abs(h[2]-h[1]);

	for(int i=3;i<=n;i++)

		dp[i]=min(dp[i-1]+abs(h[i]-h[i-1]),dp[i-2]+abs(h[i]-h[i-2]));

	cout<<dp[n]<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;

const int maxn = 2e5+2;

const int mod = 1e9+7;

const int inf = 1e9;



typedef long long ll;



int n;

int arr[maxn], dp[maxn];



int solve_dp (int pos){

    if (pos > n) return inf;

    if (pos == n - 1) return 0;



    if (dp[pos] != -1) return dp[pos];



    int ans = inf;

    ans = min(ans, solve_dp(pos + 1) + abs(arr[pos + 1] - arr[pos]));

    ans = min(ans, solve_dp(pos + 2) + abs(arr[pos + 2] - arr[pos]));



    return dp[pos] = ans;

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);



    cin >> n;



    for (int i = 0; i < n; i++){

        cin >> arr[i];

    }



    memset(dp, -1, sizeof dp);



    cout << solve_dp(0) << "\n";



    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

#define abdelrahman010 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;

using namespace std;

ll n , arr[100005] , memo[100005];

ll solve(ll i)

{

	if(i == n - 1)

		return 0;

	ll &ans = memo[i];

	if(~ans)

		return ans;

	ans = abs(arr[i] - arr[i + 1]) + solve(i + 1);

	if(i < n - 2)

		ans = min(ans , abs(arr[i] - arr[i + 2]) + solve(i + 2));

	return ans;

}

int main()

{

	abdelrahman010

	cin >> n;

	memset(memo , -1 , sizeof memo);

	for(int i = 0;i < n;i++)

		cin >> arr[i];

	cout << solve(0);

	return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

const int mod = 1e9+9;

const int nax = 1e5+3;;



int n;;

int dp[nax];

int a[nax];

int f(int i){

	if (i==n-1)

             return 0;

	if(dp[i]!=-1) 

             return dp[i];

	int ans = abs(a[i]-a[i+1])+f(i+1);

	if(i!=n-2) 

             ans = min(ans, abs(a[i]-a[i+2])+f(i+2));

	return dp[i]=ans;

}

int main(){



	cin>>n;



	for(int i=0; i<n; i++) 

               cin>>a[i];;;

	memset(dp, -1, sizeof dp);



	cout<<f(0)<<endl;;

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;





int main()

{

	int n;

	cin >> n;



	vector<int> h(n), dp(n, 1 << 30);    



	for(int &x : h)

		cin >> x;



	dp[0] = 0;

	for(int i = 1; i < n; i++)

	{

		for(int j = 1; j <= min(i, 2); j++)

			dp[i] = min(dp[i], dp[i - j] + abs(h[i] - h[i - j]));

	}



	cout << dp[n - 1] << endl;



}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

    for (int i = 0; i < n; i++)

    {

        scanf("%d", &a[i]);

    }

    dp[0] = 0;

    dp[1] = abs(a[0] - a[1]);

    for (int i = 2; i < n; i++)

    {

        dp[i] = min((dp[i - 1] + abs(a[i] - a[i - 1])), (dp[i - 2] + abs(a[i] - a[i - 2])));

    }

    printf("%d\n", dp[n - 1]);

}

// Fastest known solution for this problem: