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: 1.4402411617723483x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    cin>>n;

    int arra[n+5];

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

    {

        cin>>arra[i];



    }

    int dp[n+5];

    dp[0]=0;

    if(n>1)

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

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

    {

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

    }

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





}
// 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: 1.416483145372775x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    cin>>n;

    int arra[n+5];

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

    {

        cin>>arra[i];



    }

    int dp[n+5];

    dp[0]=0;

    if(n>1)

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

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

    {

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

    }

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





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





using namespace std;

int n;

vector<int> vec;



int meme[100007];



int solve(int index){



    if(index==n-1){

        return 0;

    }

    if(meme[index]!=-1){

        return meme[index];

    }

    

    int ret1 = 100000000;

    int ret2 = 100000000;

    if(index+1<n){

         ret1 = abs (vec[index]- vec[index+1]) + solve(index+1); 

    }

    if(index+2<n){

         ret2 = abs (vec[index]- vec[index+2]) + solve(index+2); 

    }





    return meme[index] = min(ret1,ret2);

}



int main(){

    cin>>n;

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

    int num;

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

        cin>>num;

        vec.push_back(num);

    }

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4032069187530483x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    cin>>n;

    int arra[n+5];

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

    {

        cin>>arra[i];



    }

    int dp[n+5];

    dp[0]=0;

    if(n>1)

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

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

    {

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

    }

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





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

#define ll long long

using namespace std;

#define inf 0x3f3f3f3f 

int main(){

	ll t , h[100001] , dp[100001] = {0};

    cin >> t;

	for(int i = 0 ; i < t ; ++i) scanf("%lld",h+i);

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

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

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

        cout<<dp[t-1];

}

// Faster C++ solution for the same problem (measured speedup: 1.8255960749380917x):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    cin>>n;

    int arra[n+5];

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

    {

        cin>>arra[i];



    }

    int dp[n+5];

    dp[0]=0;

    if(n>1)

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

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

    {

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

    }

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





}
// 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: 1.8273919878529525x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    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: 1.8382179776837373x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    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: 1.28260956797833x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    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: 1.2805175726000477x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    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: 1.2629028987221624x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    return 0;

}
// 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: 1.1337255090737244x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

#define gcd(a,b) __gcd(a,b)

#define lcm(a,b) (a*(b/gcd(a,b)))

#define mem(a,b) memset(a,b,sizeof(a))



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    return 0;

}