stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

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

#define fore(i,a,b) for(int i = a;i < b; i+= 1)

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

#define FL          freopen("in", "r", stdin),freopen("out", "w", stdout);

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

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

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int n, k;

int v[N];

int dp[N];

int f(int i){

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

    int &ans = dp[i];

    if(ans != -1)return ans;

    ans = 1e9;

    for(int j = 1; j <= 2; j++){

        if(i + j < n)

            ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

    cls(dp, -1);

    cout << f(0) << '\n';

    EjecuteTime;

    return 0;

}       

// Faster C++ solution for the same problem (measured speedup: 1.4403855355032773x):
#include <bits/stdc++.h>



using namespace std;



int main(){

    

    int N; cin >> N;



    int a[N], cost1=0, cost2=0, dp[N];



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

        int h; 

        cin >> h;

        a[i] = h;

    }



    dp[0] = 0;

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

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

        cost1= dp[i-1] + abs(a[i] - a[i-1]);

        cost2= dp[i-2] + abs(a[i] - a[i-2]);

        dp[i] = min(cost1, cost2);

    }

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



    return 0;

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



#define clr(x) memset((x), 0, sizeof(x))

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

#define pb push_back

#define mp make_pair

#define For(i, st, en) for(int i=(st); i<=(int)(en); i++)

#define Ford(i, st, en) for(int i=(st); i>=(int)(en); i--)

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

#define forn2(i, n) for(int i=0; i<=(int)(n); i++)

#define ford(i, n) for(int i=(n)-1; i>=0; i--)

#define fori(it, x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)

#define in(x) int (x); input((x));

#define x first

#define y second

#define less(a,b) ((a) < (b) - EPS)

#define more(a,b) ((a) > (b) + EPS)

#define eq(a,b) (fabs((a) - (b)) < EPS)

#define remax(a, b) ((a) = (b) > (a) ? (b) : (a))

#define remin(a, b) ((a) = (b) < (a) ? (b) : (a))

#define salfinal(x) cout<<x<<endl;

using namespace std;

const int MAX =1e5+1;

int v[MAX];

int memo[MAX];

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: 1.4166251375348684x):
#include <bits/stdc++.h>



using namespace std;



int main(){

    

    int N; cin >> N;



    int a[N], cost1=0, cost2=0, dp[N];



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

        int h; 

        cin >> h;

        a[i] = h;

    }



    dp[0] = 0;

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

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

        cost1= dp[i-1] + abs(a[i] - a[i-1]);

        cost2= dp[i-2] + abs(a[i] - a[i-2]);

        dp[i] = min(cost1, cost2);

    }

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



    return 0;

}
// Slower C++ solution:
#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);

}


// Faster C++ solution for the same problem (measured speedup: 2.5567088381056675x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.377936462389795x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.3981044946611205x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.3734632334414576x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 2.57185550836886x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.79450235093893x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.7915754348155783x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.3825938999676355x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.766930699212435x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.398458513120191x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.3937697238727893x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}
// 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: 1.370226871751231x):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

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

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}