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

using namespace std;

#define ll long long int

#define dbl double

#define pi acos(-1)

#define sc1(a) scanf("%lld",&a)

#define sc2(a,b) scanf("%lld%lld",&a,&b)

#define sc3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)

#define pb push_back



ll ara[1000005];



map<ll,ll>vis;

map<ll,ll>vis1;

map<ll,ll>pos;

map<ll,ll>high;

map<ll,ll>low;

map<string,ll>mp;

vector<ll>vec;

vector<ll>v;

stack<ll>q;

string str1[100];

string str2[100];

priority_queue<ll>pq;

ll dp[1000005],t;

ll mn=1000000;

ll call(ll id)

{

    if(id==t )return 0;

    if(dp[id]!=-1)

        return dp[id];

    ll p=call(id+1)+abs(ara[id+1]-ara[id]);

    ll q=100000000;

    if(id+2!=t+1)

        q=call(id+2)+abs(ara[id+2]-ara[id]);

    dp[id]=min(p,q);

    return dp[id];



}

int main()

{



    // ll t;

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

    sc1(t);

    for(ll i=1; i<=t; i++)

    {

        sc1(ara[i]);

    }

    cout<<call(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 9.281276695571696x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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



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

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

#define endl '\n'

#define de(x) cout<<#x<<" = "<<(x)<<endl

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

int dp[N];

int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int n;

	cin>>n;

	vector<int> a(n);

	for(auto& it:a) cin>>it;

	ms(dp,INF);

	dp[0]=0;

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

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

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

	}

	cout<<dp[n-1];

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4106077880791632x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

using namespace std;



int a[1000001];

int dp[1000001];

int it;



int fr(int p){

	if (p==it-1){

      return 0;

    }

	if(dp[p]!=-1){

      return dp[p];

    }

	int re=abs(a[p]-a[p+1])+fr(p+1);

	if(p!=it-2){

      re=min(re,abs(a[p]-a[p+2])+fr(p+2));

    }

	return dp[p]=re;

}

int main(){

	cin>>it;

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

    cin>>a[i];

    }



	memset(dp,-1,sizeof dp);

	cout<<fr(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.015119769865928x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

}
// Slower C++ solution:


#include <iostream>

#include <bits/stdc++.h>





using namespace std;





int a[100005], p[100005],n, i;

int respuesta(int x)

{

    if(x==n) return 0;

    if(x>n) return 1e9;

    if(p[x]!=-1)return p[x];

    return p[x]=min(abs(a[x+1]-a[x])+respuesta(x+1), abs(a[x+2]-a[x])+respuesta(x+2));

}





int main()

{

    memset(p,-1,sizeof p);

    cin>>n;

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

    {

        cin>>a[i];

    }

    cout<<respuesta(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3847928537544327x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    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: 1.3817405779724388x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    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;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1=y;

	int y1=x-(a/b)*y;

	x=x1;

	y=y1;

} 

* */

//===========================================================//

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: 1.3817405779724388x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

using namespace std;

typedef long long ll;

ll dp[100009];

int arr[100009];

ll sum = 0;

long long Frog1(int n){

 

    if(n==0) return 0;

    else if(n==1) return abs(arr[n] - arr[n-1]);

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

    else {

            ll x = Frog1(n-1);

            ll y = Frog1(n-2);

        dp[n] = min(abs (arr[n-1] - arr[n]) + x,abs(arr[n-2] - arr[n]) + y);

        //cout << n << " = " << dp[n] << endl;

         return dp[n];

    }

}

 

 

int main()

{

    int n;

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

    cin >> n;

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

    cout << Frog1(n-1) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.7783206286444944x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

using namespace std;

#define ll long long

#define ld long double

#define sf scanf

#define pf printf

#define pb push_back

#define llu unsigned long long

#define U unsigned int

#define SIZE 1000002

#define pie 3.14159265358979323

#define minuss 1e-6

#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

 

#define BIG 100000000

 

ll  lcm(ll a,ll b)

{

    ll GCD=__gcd(a,b);

    ll lcm=(a*b)/GCD;

    return lcm;

}

ll dp[100005];

ll weight[1000005],value[1000005];

ll total,val;

ll call(ll n)

{

    if(n>=total)

        return 0;

 

    if(dp[n]!=-1)

        return dp[n];

 

    ll res1=BIG,res2=BIG;

 

   // if(weight[n]+c<=val)

    res1=abs(weight[n]-weight[n+1]) + call(n+1);

    if(n+2<=total) res2=abs(weight[n]-weight[n+2]) + call(n+2);

    //cout<<n<<" "<<res1<<" "<<res2<<endl;

    dp[n]=min(res1,res2);

    return dp[n];

 

}

int main()

{

    cin>>total;

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

        cin>>weight[i];

 

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

    ll ret=call(1);

    cout<<ret<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.7776538818654075x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

using namespace std;



int num[100001], memo[100001];



int dp(int n){

    if(n==0) return 0;

    if(n==1) return abs(num[0]-num[1]);

    if(memo[n]!=-1) return memo[n];

    return memo[n]=min(dp(n-1)+abs(num[n-1]-num[n]), dp(n-2)+abs(num[n-2]-num[n]));

}



int main(){

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

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

    ios_base::sync_with_stdio(0);

    cin.tie(0);

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

    int n;

    cin>>n;

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

    cout<<dp(n-1)<<"\n";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3913143602748086x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    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()

{



    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: 1.3770358529868054x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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



using namespace std;



const int N = 1e5 + 9;



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



int solve(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]) + solve(i + 1), abs(h[i] - h[i + 2]) + solve(i + 2));

}



int main(){



    cin >> n;

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

        cin >> h[i];



    memset(dp, -1, sizeof dp);



    cout << solve(0);

}


// Faster C++ solution for the same problem (measured speedup: 1.3800372591402805x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

using namespace std;

int n;

long long h[100005];

long long dp[100005];



long long solve(int i){

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

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



	dp[i] = 1e17;

	if(i+1 <= n-1)

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

	if(i+2 <= n-1)

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



	return dp[i];

}



int main(){

	scanf("%d", &n);

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

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

	memset(dp, -1, sizeof dp);

	cout<<solve(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.778238487659303x):
#include "bits/stdc++.h"



using namespace std;



typedef pair<int, int> pii;

typedef long long ll;



#define MAX_N 100000



ll dp[MAX_N], h[MAX_N], N;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);



    cin >> N;



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

        cin >> h[i];



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



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

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

    }



    cout << dp[N - 1] << "\n";



    // --------------------

    return 0;

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

#include <cmath>

using namespace std;

int n;

int h[100001];

int f[100001];



int main(){

	cin>>n;

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

		cin>>h[i];

	}

	f[1]=0;

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

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

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

	}

	cout<<f[n]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.24607629032069x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

}
// Slower C++ solution:
/*

	Author: Luis Manuel D�az Bar�n (LUISMO)

	Problem:

	Online Judge:

	Idea:

*/

#include<bits/stdc++.h>

// Types

#define ll long long

#define ull unsigned long long

// IO

#define sf scanf

#define pf printf

#define mkp make_pair

#define fi first

#define se second

#define endl "\n"



using namespace std;



typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

const ll inf = 1e16 + 3;

const int mod = 1e9 + 7;

const int lim = 2e5 + 2;



int N;

ll hi[lim];



ll dp[lim];

int mark[lim];



ll DynamicProgramming(int idx)

{

    if(idx <= 0)

        return 0;



    if(mark[idx])

        return dp[idx];



    mark[idx] = true;



    DynamicProgramming(idx - 1);

    DynamicProgramming(idx - 2);



    ll k1 = dp[idx-1] + abs(hi[idx] - hi[idx-1]);

    ll k2 = (idx > 1) ? dp[idx-2] + abs(hi[idx] - hi[idx-2]) : inf;



    dp[idx] = min(k1, k2);

    return dp[idx];

}





void solve()

{

    cin >> N;

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

        cin >> hi[i];   



    ll answ = DynamicProgramming(N - 1);

    cout << answ;

}



void fastIO()

{

	cin.sync_with_stdio(false);

	cin.tie(0);

}



void IO()

{

	if(fopen("d:\\lmo.in","r") != NULL)

	{

		freopen("d:\\lmo.in","r",stdin);

	}

	else if(fopen("/media/Beijing/lmo.in","r") != NULL)

	{

		freopen("/media/Beijing/lmo.in", "r", stdin);

	}

}



int main()

{

	IO();



	fastIO();



	solve();

}


// Faster C++ solution for the same problem (measured speedup: 5.317322169624953x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

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

#define ll long long

#define ull unsigned long long

using namespace std;

ll n,a[100009],dp[100009];

	

int main (){

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

	cin >> n;

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

		cin >> a[i];

	}

	for (int i=n; i>=1; i--){

		ll x,y;

		if (i<=n-2){

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

		}

		else if (i==n-1) dp[i]=abs(a[i]-a[i+1])+dp[i+1];

	}

	cout << dp[1] << endl;	

}

// Faster C++ solution for the same problem (measured speedup: 5.3088577452842864x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

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

#include<stdio.h>



using namespace std;



int main(){

    int n, i, j;

    scanf("%d", &n);

    int height[n+5];

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

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

    }

    int dp[n+5];

    for(i = 0; i <=n; i++) dp[i] = 1e9;

    dp[1] = 0;

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

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

        if(i > 2) {

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

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

            }

        }

    }



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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.1801357856349695x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

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

#include <iostream>

#include <string>

#include <vector>

#include <cmath>

#include <algorithm>

#include <sstream>

#include <queue>

#include <deque>

#include <set>

#include <bitset>

#include <iterator>

#include <list>

#include <stack>

#include <map>

#include <set>

#include <functional>

#include <numeric>

#include <utility>



#include <cstdlib>  // for abs



#include <iomanip>



using namespace std;

int main()

{

  // Standard incantation to set up fast IO.

  ios::sync_with_stdio(false);

  cin.tie(0);



  int n;

  cin >> n;

  int cur, next;

  if (n <= 1) {

    cout << 0 << '\n';

    return 0;

  }

  int a, b, c;

  cin >> a >> b;

  cur = 0;

  next = abs(a - b);

  if (n == 2) {

    cout << next << '\n';

    return 0;

  }

  int ans;

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

    cin >> c;

    ans = min(cur + abs(c-a), next + abs(c-b));

    a = b;

    b = c;

    cur = next;

    next = ans;

  }

  cout << ans << '\n';



  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.311763486948631x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

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

using namespace std;



int main(){



        int n; cin >> n;

        vector<int> frog(n + 1, 0);

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

                cin >> frog[i];

        }



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

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



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

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

        }



        cout << dp[n] << endl;









        return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.295489360385033x):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

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

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

    }

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

    return 0;

}