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

    #define maxn 100000

    #define INF 1LL<<60

    using namespace std;

    long long h[maxn+3], dp[maxn+3];

    int n;



    long long solve(int cur){

        if(cur >= n) return INF;

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

        if(dp[cur]!= INF) return dp[cur];

        //long long res1 = INF, res2 = INF;

        long long res1 = solve(cur+1) + llabs(h[cur+1]-h[cur]);

        long long res2 = solve(cur+2) + llabs(h[cur+2]-h[cur]);

        return dp[cur] = min(res1, res2); 

    }



    int main() {

    	// your code goes here

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

    	while(scanf("%d",&n)==1){

            

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

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

                dp[i] = INF;

            }



            long long ans = solve(0);

            printf("%lld\n", ans);

        }

    }

     

// Faster C++ solution for the same problem (measured speedup: 1.6729375962811055x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	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: 3.076805845473983x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include<cmath>

#include<cstdio>

#include<cstring>

using namespace std;

int n,m,f[1100000],w[1100000];

int main ()

{

	cin>>n;

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

	{

		cin>>w[i];

	}

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

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

	{

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

	}	

	cout<<f[n];

}

// Faster C++ solution for the same problem (measured speedup: 1.6903222671684048x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include <algorithm>

#include <cmath>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <iomanip>

#include <string>

#include <set>

#include <vector>

#include <queue>

#include <stack>

#include <map>

using namespace std;

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

#define fore(i, a, n) for (long long i = a; i < n; i++)

#define ford(i, a, n) for (long long i = n - 1; i >= a; i--)

#define si(a) scanf("%d", &a)

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

#define sii(a, b) scanf("%d%d", &a, &b)

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

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

#define debug(a) cout << a << endl

#define pr(a) printf("%d ",a)

#define endl '\n'

#define pi acos(-1.0)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define IO ios::sync_with_stdio(false), cin.tie(0)

#define ull unsigned long long

const double eps = 1e-8;

inline int sgn(const double &x) { return x < -eps ? -1 : x > eps; }

typedef long long ll;

const int inf=0x3f3f3f3f;

const int MAXN = 1e5 + 10;

int n,a[MAXN];

ll dp[MAXN];

int main()

{

	si(n);

	for(int i=1;i<=n;++i)si(a[i]);

	dp[1]=0;dp[2]=abs(a[1]-a[2]);

	for(int i=3;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("%lld\n",dp[n]);

 } 

// Faster C++ solution for the same problem (measured speedup: 1.6715111472184243x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	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: 3.071787433192256x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

using namespace std;

#define ll long long

#define pb push_back

#define F first

#define S second

const int mod = 1e9+7, A = 1e5+5;

ll n, k, h[A], dp[A];

ll table1()

{

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

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

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

	return dp[n-1];

}

ll table2()

{

	dp[0] = 0;

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

	{

		dp[i] = 1e9;

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

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

	}

	return dp[n-1];

}

int main()

{

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

	cin >> n;

	//cin >> k;

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

	cout << table1();

	//cout << table2();

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7117466921863347x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

	printf("%d\n", dp[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: 2.3705586070206612x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include<vector>

using namespace std;

 

int dp[100010];

int n;

int conti;

bool pasa;

int main(){

     cin >> n;

    vector<int> h(n);

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

        conti ++;

        cin >> h[i];

        if(conti > 0){

        conti--;

        pasa = true;

       //printf("prueba")

}

    }

    dp[0] = 0;

    dp[1] = abs(h[1] - h[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] << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.7089143073520823x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include <vector>

#include <cmath>

#include <map>

#include<algorithm>

#include<string.h>

using namespace std;

int main() {

	//freopen("1.txt","r",stdin);

	//freopen("2.txt","w",stdout); 

   	int n;

   	cin>>n;

   	int a[100000];

   	int b[100000];

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

   		cin>>a[i];

   	b[0]=0;

   	b[1]=abs(a[1]-a[0]);

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

   	{

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

	}

	cout<<b[n-1];

}

// Faster C++ solution for the same problem (measured speedup: 1.7158955337344846x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include<cstdio>

#include<cmath>

#include<cstring>

#define Max(a,b) (a>b?a:b)

#define Min(a,b) (a<b?a:b)

using namespace std;

const int MAXN = 1e5+10, INF = 0x3f3f3f3f;

int A[MAXN], dp[MAXN], N;



int main()

{

	scanf("%d", &N);

	int i;

	for(i=1;i<=N;i++) scanf("%d", &A[i]), dp[i] = INF;

	dp[1] = 0;

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

	for(i=3;i<=N;i++) {

		dp[i] = Min(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;

}

// Faster C++ solution for the same problem (measured speedup: 1.684057669041217x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	return 0;

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

#include <vector>

#include <algorithm>

#include <stdio.h>

#include <map>

using namespace std;

#define min(a,b) (a>b?b:a);

int main(){

	int n;

	cin>>n;

	vector<int> path(n);

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

	vector<int> dp(n);

	dp[0]=0;

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

	if(n==2){

		cout<<dp[1]<<endl;

		return 0;

	}

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

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

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6957400406232284x):
#include<stdio.h>

#include<cmath>

#include<algorithm>

using namespace std;

const int maxn = 100005;

const int INF = 1000000000;



int A[maxn];//石头高度

int dp[maxn] = {0}; 

int n;



int main(){

//	freopen("4.txt","r",stdin);

	scanf("%d",&n);

	fill(dp,dp+maxn,INF);

	

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

		scanf("%d\n",&A[i]);

	}

	dp[1] = 0;

	for(int i=2;i<=n;i++){//跳到i上 

		int j1 = i-2, j2 = i-1;

		if(j1>=1){//

			dp[i] = min(dp[i] , dp[j1] + abs(A[i]-A[j1]));

		}

		if(j2>=1){//

			dp[i] = min(dp[i] , dp[j2] + abs(A[i]-A[j2]));

		}

	}

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

	

	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: 1.4469774277198626x):
#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;

}
// 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: 1.4555497335097525x):
#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;

}
// 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: 2.5903728923423466x):
#include<iostream>

#include<algorithm>

#define MAXN 100010 

using namespace std;

int h[MAXN];

int dp[MAXN];

int main() {

	int N;

	cin >> N;

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

		cin >> h[i];

	}

	dp[0] = 0;

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

	for (int i = 2; i <= N-1; 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] << endl;

	//system("pause");

	return 0;

}
// Slower C++ solution:
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

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

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

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

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

#define clr(v,d)                memset(v , d , sizeof(v))

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

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

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

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

        if(j>=2){

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

        }

        else if(j ==1){

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

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

//    cout << solve(0, 1)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.8151507051306437x):
#include<iostream>

#include<algorithm>

#define MAXN 100010 

using namespace std;

int h[MAXN];

int dp[MAXN];

int main() {

	int N;

	cin >> N;

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

		cin >> h[i];

	}

	dp[0] = 0;

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

	for (int i = 2; i <= N-1; 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] << endl;

	//system("pause");

	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: 1.3960796811690428x):
#include<iostream>

#include<algorithm>

#define MAXN 100010 

using namespace std;

int h[MAXN];

int dp[MAXN];

int main() {

	int N;

	cin >> N;

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

		cin >> h[i];

	}

	dp[0] = 0;

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

	for (int i = 2; i <= N-1; 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] << endl;

	//system("pause");

	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: 1.416513264887647x):
#include<iostream>

#include<algorithm>

#define MAXN 100010 

using namespace std;

int h[MAXN];

int dp[MAXN];

int main() {

	int N;

	cin >> N;

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

		cin >> h[i];

	}

	dp[0] = 0;

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

	for (int i = 2; i <= N-1; 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] << endl;

	//system("pause");

	return 0;

}