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

#define pii pair<int,int>

#define vi vector<int>

#define ppii pair<pii>



using namespace std;

const int maxn=1e5+5;



int n;

vi arr;

int dp[maxn];



int dfs(int pos)

{

    if(pos==n){return 0;}



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



    int ans=(1<<29);



    if(pos<n)

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

    if(pos<n-1)

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

    dp[pos]=ans;



    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

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



    cin>>n;



    arr=vi(n+1);



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

        cin>>arr[i];



    memset(dp,-1,sizeof dp);



    cout<<dfs(1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3798753011918208x):
#include <bits/stdc++.h>

#include<vector>

#include<stdlib.h>

#include<math.h>



using namespace std;



int main() {

    long long n, x, t;

    vector<long long> ara;

    long long arr[100007];

    cin >> n;

    t = n;

    while(n--){

        cin >> x;

        ara.push_back(x);

    }

    arr[0] = 0;

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

        i==1? arr[i] = abs(ara[i]-ara[i-1]) : arr[i] = min(arr[i-2]+abs(ara[i]-ara[i-2]), arr[i-1]+abs(ara[i]-ara[i-1]));

    }

    //for(int j = 0; j<t; j++) printf("%lld %lld\n", ara[j], arr[j]);

    cout << arr[t-1] << endl;

    return 0;

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

#include <queue>

typedef long long ll;



using namespace std;



const int N=1e5+5;





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

   int n,k=2;

   cin>>n;

   int arr[N]={},dp[N]={};

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

    cin>>arr[i];

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

   dp[0]=0;

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

   {



     for(int j=1;j<=k;j++)

     {

         if(i-j>=0)

         {

             if(dp[i]==-1)

             {

                 dp[i]=dp[i-j]+abs(arr[i]-arr[i-j]);

             }

             else

             {

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





             }





         }



     }









   }

   cout<<dp[n-1];



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.816500520066382x):
#include <bits/stdc++.h>

#include<vector>

#include<stdlib.h>

#include<math.h>



using namespace std;



int main() {

    long long n, x, t;

    vector<long long> ara;

    long long arr[100007];

    cin >> n;

    t = n;

    while(n--){

        cin >> x;

        ara.push_back(x);

    }

    arr[0] = 0;

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

        i==1? arr[i] = abs(ara[i]-ara[i-1]) : arr[i] = min(arr[i-2]+abs(ara[i]-ara[i-2]), arr[i-1]+abs(ara[i]-ara[i-1]));

    }

    //for(int j = 0; j<t; j++) printf("%lld %lld\n", ara[j], arr[j]);

    cout << arr[t-1] << endl;

    return 0;

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

using namespace std;



typedef long long ll;

#define mp make_pair

#define f first

#define s second

#define pb push_back



const int inf = 1e7;

const int N  = 1e5 + 5;



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

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



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



int solve(int idx) {



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



  int &ret = dp[idx];



  if(ret!=-1) return ret;



  ret = solve(idx+1) + abs(h[idx]-h[idx+1]);



  if(idx!=n-2)

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



  return ret;



}



int main() {



    ios_base::sync_with_stdio(false);

    cin.tie(0);



    cin >> n;



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



      cin >> h[i];



    }



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



    cout << solve(0) << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3789035481205154x):
#include <bits/stdc++.h>

#include<vector>

#include<stdlib.h>

#include<math.h>



using namespace std;



int main() {

    long long n, x, t;

    vector<long long> ara;

    long long arr[100007];

    cin >> n;

    t = n;

    while(n--){

        cin >> x;

        ara.push_back(x);

    }

    arr[0] = 0;

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

        i==1? arr[i] = abs(ara[i]-ara[i-1]) : arr[i] = min(arr[i-2]+abs(ara[i]-ara[i-2]), arr[i-1]+abs(ara[i]-ara[i-1]));

    }

    //for(int j = 0; j<t; j++) printf("%lld %lld\n", ara[j], arr[j]);

    cout << arr[t-1] << endl;

    return 0;

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

using namespace std;



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: 1.365918352676421x):
#include <bits/stdc++.h>

#include<vector>

#include<stdlib.h>

#include<math.h>



using namespace std;



int main() {

    long long n, x, t;

    vector<long long> ara;

    long long arr[100007];

    cin >> n;

    t = n;

    while(n--){

        cin >> x;

        ara.push_back(x);

    }

    arr[0] = 0;

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

        i==1? arr[i] = abs(ara[i]-ara[i-1]) : arr[i] = min(arr[i-2]+abs(ara[i]-ara[i-2]), arr[i-1]+abs(ara[i]-ara[i-1]));

    }

    //for(int j = 0; j<t; j++) printf("%lld %lld\n", ara[j], arr[j]);

    cout << arr[t-1] << endl;

    return 0;

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

using namespace std;

int a,b,c,d,dp[100000],cost[100000],weight[100];



void func(int a)

{

    dp[1]=0;

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

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

    {

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

    }

}



int main()

{

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

    cin>>a;

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

    {

        cin>>cost[i];

    }

    func(a);

    cout<<dp[a]<<endl;

    //cout<<nCr(a,b)<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 1.3613511645974563x):
#include <bits/stdc++.h>

#include<vector>

#include<stdlib.h>

#include<math.h>



using namespace std;



int main() {

    long long n, x, t;

    vector<long long> ara;

    long long arr[100007];

    cin >> n;

    t = n;

    while(n--){

        cin >> x;

        ara.push_back(x);

    }

    arr[0] = 0;

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

        i==1? arr[i] = abs(ara[i]-ara[i-1]) : arr[i] = min(arr[i-2]+abs(ara[i]-ara[i-2]), arr[i-1]+abs(ara[i]-ara[i-1]));

    }

    //for(int j = 0; j<t; j++) printf("%lld %lld\n", ara[j], arr[j]);

    cout << arr[t-1] << endl;

    return 0;

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

using namespace std;

#define    sz    100110

#define    ll    long long int

ll n,c=0,p,q;

ll num[sz];

ll vis[sz];

ll dp[sz];



ll tani(int pos)

{



  if(pos>=n)return 0;

  if(vis[pos]==1)return dp[pos];

  vis[pos]=1;



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



}



int main()

{

    ll t,i;



    for(i=1;i<sz;i++)num[i]=1000000;



    cin>>n;

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

    {

        cin>>num[i];

    }

    t=tani(1);

    cout<<t<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8005212091595229x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;



#define fr(i,n) _back

#define pb	push_back

#define eb	emplace_back

#define mk	make_pair

#define fi	first

#define se	second

#define endl	'\n'



typedef long long ll;

typedef pair<int,int> ii;

typedef vector<ii> vii;

const int INF = 0x3f3f3f3f;

const double PI = acos(-1.0);

const int N = 1e5 + 3;

int h[N];

int dp[N];



int main() {

    ios_base::sync_with_stdio(false);

    memset(dp, INF, sizeof dp);

    int n; cin >> n;

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



    dp[0] = 0;

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

        for(int j = max(0,i-2); j < i; j++)

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



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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4156249901082547x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;



int N, a[100100], dp[100100];



int f(int n) {

	if (n==1) return 0;

	if (n==2) return abs(a[2] - a[1]);

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

	dp[n] = min(

				f(n-1) + abs(a[n] - a[n-1]),

				f(n-2) + abs(a[n] - a[n-2])

				);

	return dp[n];

}



int main() {

	cin >> N;

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

		cin >> a[i];

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

	cout << f(N) << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3997972420105527x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;

typedef long long LL;



int arr[100007];

int dp[100007];



int Solve(int idx, int n){

//	cout<<cost<<" "<<idx<<endl;

	if(idx >= n)return 1e9;

	if(idx == n - 1){

		return 0;	

	}

	

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

	

	

	return dp[idx] = min(abs(arr[idx] - arr[idx+1]) + Solve(idx+1, n), abs(arr[idx] - arr[idx+2]) + Solve(idx+2,n));

}



int main(){

	int n; cin>>n;

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

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

		

	cout<<Solve(0,n)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4003928824667795x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;

int dp[100005];

int H[100005];

void memo(int i,int n)

{

	if(i==n) return;

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

	memo(i+1,n);

}

int main()

{

	int N;cin>>N;

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

	{

        cin>>H[i];

	}

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

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

	memo(2,N);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3996877861164225x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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



using namespace std;



int dp[(int)1e5+5];

int main()

{



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







    fill(dp, dp+(int)1e5+5 , INT_MAX );



    dp[1] = 0;

    int n;

    cin>>n;

    vector <int> v(n+3,0);

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



        cin>>v[i];



    }

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



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

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



    }

    cout<<dp[n]<<"\n";

}


// Faster C++ solution for the same problem (measured speedup: 1.4188105864611047x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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



using namespace std;



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

#define DB( x ) cout << #x << " = " << x << endl;

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

#define endl '\n'



typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll, ll> pii;

typedef pair<int, pii> tri;

typedef vector <vector <ll>> matrix;

typedef pair<string, string> pss;



const ll mod = (ll)1e9 + 7;

const int MAX = (int)1e6 + 100;



const ll oo = 0x3f3f3f3f3f3f3f3f;



ll dp[MAX], a[MAX];



int main() {

    ios_base::sync_with_stdio(0);

    cin.tie(0);



    memset(dp, oo, sizeof(dp));



    int n; cin >> n;



    dp[1] = 0;



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

        cin >> a[i];

    }



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

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

        if(i - 2 >= 1){

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

        }

    }



    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 9.40447836418828x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



*/
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int h[N];

int mem[N];



int go(int idx){

    if(idx >= n)

        return 0;

    if(mem[idx] != -1)

        return mem[idx];

    int best = 1e9;

    if(idx + 1 <= n)

        best = go(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 <= n)

        best = min(best , go(idx + 2)  + abs(h[idx] - h[idx + 2]));

    return mem[idx] = best;

}



int main(){

    scanf("%d" , &n);

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

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

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1));

}

//1 2 3


// Faster C++ solution for the same problem (measured speedup: 1.3872995154507388x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

#define pii pair<int,int>

#define vi vector<int>

#define ppii pair<pii>



using namespace std;

const int maxn=1e5+5;



int n;

vi arr;

int dp[maxn];



int dfs(int pos)

{

    if(pos==n){return 0;}



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



    int ans=(1<<29);



    if(pos<n)

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

    if(pos<n-1)

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

    dp[pos]=ans;



    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

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



    cin>>n;



    arr=vi(n+1);



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

        cin>>arr[i];



    memset(dp,-1,sizeof dp);



    cout<<dfs(1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4138794278126217x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

#include <queue>

typedef long long ll;



using namespace std;



const int N=1e5+5;





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

   int n,k=2;

   cin>>n;

   int arr[N]={},dp[N]={};

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

    cin>>arr[i];

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

   dp[0]=0;

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

   {



     for(int j=1;j<=k;j++)

     {

         if(i-j>=0)

         {

             if(dp[i]==-1)

             {

                 dp[i]=dp[i-j]+abs(arr[i]-arr[i-j]);

             }

             else

             {

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





             }





         }



     }









   }

   cout<<dp[n-1];



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8612643575216488x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;



typedef long long ll;

#define mp make_pair

#define f first

#define s second

#define pb push_back



const int inf = 1e7;

const int N  = 1e5 + 5;



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

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



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



int solve(int idx) {



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



  int &ret = dp[idx];



  if(ret!=-1) return ret;



  ret = solve(idx+1) + abs(h[idx]-h[idx+1]);



  if(idx!=n-2)

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



  return ret;



}



int main() {



    ios_base::sync_with_stdio(false);

    cin.tie(0);



    cin >> n;



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



      cin >> h[i];



    }



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



    cout << solve(0) << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.412883727929273x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;



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: 1.3995785396353206x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;

int a,b,c,d,dp[100000],cost[100000],weight[100];



void func(int a)

{

    dp[1]=0;

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

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

    {

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

    }

}



int main()

{

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

    cin>>a;

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

    {

        cin>>cost[i];

    }

    func(a);

    cout<<dp[a]<<endl;

    //cout<<nCr(a,b)<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 1.3948988028053173x):
#include<iostream>

#include<cmath> 

using namespace std;



const int maxN = 100015;

int f[maxN];//f[i]表示跳到第只石头上需要的最小能量

int n;

int arr[maxN];



int main(){

	cin >> n;

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

		cin >> arr[i];

	} 

	

	f[0] = 0, f[1] = 0;//起初在第一只石头上 

	f[2] = abs(arr[1]-arr[2]);//	

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

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

	}

	cout << f[n]<<"\n";

} 

/*



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

using namespace std;

#define    sz    100110

#define    ll    long long int

ll n,c=0,p,q;

ll num[sz];

ll vis[sz];

ll dp[sz];



ll tani(int pos)

{



  if(pos>=n)return 0;

  if(vis[pos]==1)return dp[pos];

  vis[pos]=1;



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



}



int main()

{

    ll t,i;



    for(i=1;i<sz;i++)num[i]=1000000;



    cin>>n;

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

    {

        cin>>num[i];

    }

    t=tani(1);

    cout<<t<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8245719867218275x):
#include<iostream>

#include<algorithm>

using namespace std;

int a[100010];

int dp[100010];

int main()

{

    int n;

    scanf("%d", &n);

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

    {

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

        dp[i] = 100000;

    }

    dp[1] = 0;

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

    dp[3] = abs(a[1] - a[3]);

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

    {

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

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

    }

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

    return 0;

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

using namespace std;



#define fr(i,n) _back

#define pb	push_back

#define eb	emplace_back

#define mk	make_pair

#define fi	first

#define se	second

#define endl	'\n'



typedef long long ll;

typedef pair<int,int> ii;

typedef vector<ii> vii;

const int INF = 0x3f3f3f3f;

const double PI = acos(-1.0);

const int N = 1e5 + 3;

int h[N];

int dp[N];



int main() {

    ios_base::sync_with_stdio(false);

    memset(dp, INF, sizeof dp);

    int n; cin >> n;

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



    dp[0] = 0;

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

        for(int j = max(0,i-2); j < i; j++)

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



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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4345344489780152x):
#include<iostream>

#include<algorithm>

using namespace std;

int a[100010];

int dp[100010];

int main()

{

    int n;

    scanf("%d", &n);

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

    {

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

        dp[i] = 100000;

    }

    dp[1] = 0;

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

    dp[3] = abs(a[1] - a[3]);

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

    {

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

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

    }

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

    return 0;

}