stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
// 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;

}

// Faster C++ solution for the same problem (measured speedup: 1.8161354812430626x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.39683709812465x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.4172817677024507x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}






// Faster C++ solution for the same problem (measured speedup: 1.3923025115787613x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.607132682210429x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.819116864152126x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.8161498006221386x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 1.4015584200204685x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include<iostream>

#include<stdio.h>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define maxn 100005

int dp[maxn];

int h[maxn];

int n;

int main(){

    scanf("%d",&n);

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

    memset(h,0,sizeof(h));

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

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

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

        if(i-1>=1&&i-2>=1)

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7911670224581582x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

int v[100010];

int memo[100010];

int c;



int f(int i)

{



    if(memo[i]!=-1)

        return memo[i];



    if(i == 0)

        return memo[0] = 0;



    if(i == 1)

        return memo[1] = abs(v[0] - v[1]);



    return memo[i] = min(abs(v[i] - v[i-1]) + f(i-1), abs(v[i] - v[i-2]) + f(i-2));

}



int main()

{

    ios_base::sync_with_stdio(false);

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

    cin>>c;



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

    {

        cin>>v[i];

    }



    cout<<f(c-1)<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4176406420994552x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define Me3za() ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);srand(time(NULL))

#define endl "\n"

int n;

int arr[100005];

int mem[100005];

long long solve(int i,int cur){

if(i>=n)return 1e9;

//cout<<i<<" "<<cur<<endl;

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

long long ch1=solve(i+1,arr[i+1])+abs(cur-arr[i+1]);

long long ch2=0;

if(i+2<=n)ch2=solve(i+2,arr[i+2])+abs(cur-arr[i+2]);

return mem[i]=min(ch1,ch2);

}

int main() {

Me3za();

cin>>n;

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

    cin>>arr[i];

}

memset(mem,-1,sizeof mem);

cout<<solve(0,arr[0]);

}




// Faster C++ solution for the same problem (measured speedup: 1.4128875384950266x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



int const N=1e5+5, M=1e9+7, OO = 0x3f3f3f3f;



int n;

int arr[N];



int mem[N];

int solve(int i)

{

    if(i==n-1)

        return 0;

    if(mem[i]!=-1)

        return mem[i];

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=(i<=n-3) ? solve(i+2)+abs(arr[i]-arr[i+2]) : OO;

    return mem[i]=min(op1,op2);

}





int main()

{

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

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

    scanf("%d",&n);

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

        scanf("%d",arr+i);

    memset(mem, -1, sizeof mem);

    printf("%d\n",solve(0));



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3890217579335478x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}



const int N = 1e5 + 9;

int n, h[N], dp[N];



int sol(int i){

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

    if(i == n)  return 1e9;



    int &ret = dp[i];

    if(~ret)    return ret;



    return ret = min(abs(h[i] - h[i + 1]) + sol(i + 1), abs(h[i] - h[i + 2]) + sol(i + 2));

}





int main() {



    cin >> n;



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



    memset(dp, -1, sizeof dp);



    cout << sol(0);

}

// Faster C++ solution for the same problem (measured speedup: 1.3996541523126274x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include<bits/stdc++.h>



using namespace std;



const int N=1e5;

int n;

int arr[N];

int mem[N];



int solve(int i){

    if(i==n-1)

        return 0;

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

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=i+2<=n-1?solve(i+2)+abs(arr[i]-arr[i+2]):1000000000;

    return mem[i]=min(op1,op2);

}



int main() {

    scanf("%d",&n);

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

        scanf("%d",arr+i);

    memset(mem,-1,sizeof mem);

    printf("%d",solve(0));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.388679813527163x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



// يلعن ابو الي يعقدها على نفسه تاني



int main()

{

    ios_base::sync_with_stdio(false);    cin.tie(NULL);



    ll n; cin >> n;

    ll arr[100005];

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

        cin >> arr[i];

    }

    if(n == 2){

        cout << abs(arr[1]-arr[0]);

    }else{

        ll sol[100005] = {};

        sol[1] = abs(arr[1]-arr[0]);

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

            sol[i] = min(sol[i-1]+abs(arr[i]-arr[i-1]),sol[i-2]+abs(arr[i]-arr[i-2]));

        }









        cout << sol[n-1];

    }















    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6079555652514128x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>

 

using namespace std;

 

#define sf(n) scanf("%d", &n)

#define pd(n) printf("%.12lf", n)

#define pf(n) printf("%d", n)

#define p() printf("\n")

#define ps() printf(" ")

#define ll long long

#define ull unsigned long long

#define MAX 1e9

const int mod = 1000 * 1000 * 1000 + 7;

#define ex 1e18

#define dbl 2e09

#define qu(q) queue<int> q

#define pqu(q) priority_queue<int> q

#define len(s) s.length()

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

#define pb(x) push_back(x)

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

#define rall(x) x.rbegin(), x.rend()

#define F first

#define S second

#define clr(x) x.clear()

#define LOG(x) std::cout << x << std::endl;

 

const int INF = 0x3f3f3f3f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const double PI = 3.14159265359;

 

int dx[6] = {-1,-1,0,0,1,1};

int dy[6] = {-1,0,-1,1,0,1};



int a[100005];

int dp[100005];

int n;

int solve(int idx){

    if(idx==(n-1))

        return 0;

    if(idx>=n)

        return INF;

    int &ret = dp[idx];

    if(~ret)

        return ret;

    ret = INF;

    ret=min(ret,solve(idx+1)+abs(a[idx+1]-a[idx]));

    ret=min(ret,solve(idx+2)+abs(a[idx+2]-a[idx]));

    return ret;

}







int main(){



    cin>>n;

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

        cin>>a[i];

    }

    memset(dp,-1,sizeof dp);

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4013673184160105x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int abssub(int a,int b){

	if(a>b)return a-b;

	return b-a;

}

int main(){

	int n,h[100001],bp[100001]={0};

	cin>>n; 

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

		cin>>h[i];

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

		if(i==2)bp[i]=bp[i-1]+abssub(h[i],h[i-1]);

		else{

			bp[i]=min(bp[i-1]+abssub(h[i],h[i-1]),bp[i-2]+abssub(h[i],h[i-2]));

		}

	}

	cout<<bp[n]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4125779000704253x):
# include <bits/stdc++.h>

# define int long long

# define fi first

# define se second



using namespace std ;



const int N = 1e5 + 1 ;

int v[N], dp[N];



int32_t main() {

    //freopen("haybales.in", "r", stdin ) ;

    //freopen("haybales.out", "w", stdout ) ;

    int t ;

    cin >> t ;

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

        cin >> v[i] ;

    }

    dp[2] += abs(v[2] - v[1] ) ;

    for( int i = 3 ; i <= t ; i ++ ) {

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

    }

    cout << dp[t] << endl;



}