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



using namespace std;

const int N = 1e4 + 5;



int main()

{

	int n;

	cin >> n;

	int height[n+1], dp[n+1];

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

		cin >> height[i];

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

	dp[0] = 0;

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

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

	{

		long long a, b;

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

		b = abs(height[i] - height[i - 2]) + dp[i - 2];

		dp[i] = min(a, b);

	}

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



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.671758543435203x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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



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

#define _(x) {cout << #x << " = " << x << " ";}

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

#define pb push_back

#define F first

#define S second

#define MOD 1000000007



typedef long long ll;

typedef unsigned long long ull;



const double E = 1e-8;

const double PI = acos(-1);

    

using namespace std;



int N, h[100005], dp[100005]; 



int solve() {

    dp[0] = 0;

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

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

        if(i > 1)

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

    }

    return dp[N-1];

}

    

int main(int argc, char** argv) {

    

    IOBOOST

    

    cin>>N;

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

        cin>>h[i];

    cout<<solve()<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.729993761239497x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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

#define maxn 100005

using namespace std;



int main(){

    int n;

    cin >> n;

    int h[maxn];

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

        cin >> h[i];

    }

    int dp[maxn];

    fill(dp, dp+maxn, 0x7fffffff);

    dp[0] = 0;

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

        int temp = 0;

        if(i+1 < n){

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

            if(temp < dp[i+1])

                dp[i+1] = temp;

        }

        if(i+2 < n){

            temp = abs(h[i+2]-h[i]) + dp[i];

            if(temp < dp[i+2])

                dp[i+2] = temp;

        }

    }

    cout << dp[n-1];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 6.561556196801219x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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



using namespace std;

int dp[100005],h[100005];

int main()

{

    //freopen("C://Code//input.txt","r",stdin);

    int n;scanf("%d",&n);

    for(int i=0;i<n;++i)scanf("%d",h+i);

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

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

        scanf("%d",h+i);

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

    }

    printf("%d\n",dp[n-1]);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.61297775082937x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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

using namespace std;

const int maxn=1e5+10;

int n,h[maxn],f[maxn];

int main(){

	scanf("%d",&n);

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

		scanf("%d",&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-1]-h[i]),f[i-2]+abs(h[i-2]-h[i]));

	}

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

} 

// Faster C++ solution for the same problem (measured speedup: 4.645888586078901x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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

#define inf 10000

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+3];

    int dp[100001];



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

    {

        cin>>a[i];

        dp[i]=inf;

    }

    dp[0]=0;

    dp[1]=0;

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

    for(int i=3;i<n+1;i++)

    {

        int l1=dp[i-1]+abs(a[i-2]-a[i-1]);

        int l2=dp[i-2]+abs(a[i-3]-a[i-1]);



        dp[i]=min(l1,l2);

    }



    cout<<dp[n];





}




// Faster C++ solution for the same problem (measured speedup: 4.6985145392573635x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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

using namespace std;



int main()

{

    long long int  n;

    cin>>n;

    long long int  hi[n];



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

    {

        cin>>hi[i];

    }

    long long int  ans[n];

    ans[0] = 0;

    ans[1] = abs(hi[1] - hi[0]);

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

    {

        ans[i] =  min(ans[i - 2]   + abs(hi[i] - hi[i - 2]),ans[i - 1] + abs(hi[i] -hi[i - 1]));



    }

    cout<<ans[n-1];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.659522399011774x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

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

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

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

#define fi first

#define se second

#define pp push_back

#define pb pop_back

#define pi 3.141592654

#define MOD 1000000007 

#define cons1 1e12

#define cons2 1e6

using namespace std;

typedef long long ll;

typedef vector<int> vi;

typedef pair<int, int> pa;

typedef unsigned long long int ull;

typedef double long dl;

const int N = 1e5+9;

int n;

int v[N], ans[N];

int sol(int i) {

	if (ans[i] == -1) {

		if (i == n - 1)

			ans[i] = 0;

		else if (i == n - 2)

			ans[i] = abs(v[i] - v[i + 1]);

		else 

			ans[i] = min(abs(v[i] - v[i + 1])+sol(i + 1)

				, abs(v[i] - v[i + 2]) + sol(i + 2));

		

	}

	return ans[i];

	

}

void solve() {

	clr(ans, -1);

	cin >> n;

	rep(i, 0, n)

		cin >> v[i];

	cout << sol(0) << endl;

}

int main() {

	ios::sync_with_stdio(0);

	cin.tie(NULL);

	cout.tie(NULL);

	int t = 1;

	//cin >> t;

	while (t--)

		solve();



#ifndef ONLINE_JUDGE

	- system("pause");

#endif



}

// Faster C++ solution for the same problem (measured speedup: 6.596057004011578x):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

        scanf("%d",h+i);

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

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

    }

    printf("%d\n",p[0]);



    return 0;

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

using namespace std;

int dif(int a, int b)

{

    int ans=a-b;

    if(ans<0) ans*=-1;

    return ans;

}

int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    int N, h[100005],i,j;

    int f1,f2;

    cin>>N;

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

        cin>>h[i];

    int M[100005]={0};

    M[N-2]=dif(h[N-1],h[N-2]);

    for(i=N-3;i>=0;i--)

    {

        f1=dif(h[i+1],h[i])+M[i+1];

        f2=dif(h[i+2],h[i])+M[i+2];

        M[i]=min(f1,f2);

    }

    cout<<M[0]<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4186372974702544x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

int n;

int h[1000005], memo[1000005];

int f(int n) {

    if(n == 1) {                        // bisa juga base case nya n = 0, f(0) harus tinggi, misal 1e9;

        return 0;

    }

    if(n == 2) {

        return abs(h[1] - h[2]);

    } 

    else if(memo[n] != -1) {

        return memo[n];

    }

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

    return memo[n];

}



int main() {

    scanf("%d", &n);

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

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

    }

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

    f(n);

    printf("%d\n", f(n));

}

// Faster C++ solution for the same problem (measured speedup: 5.059105883660197x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;



typedef signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX], dp[MX];



int solve(int pos) {

    if(pos > n) return INF;

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

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

    int ans = INF;

    if(pos + 1 < n) ans = min(ans, abs(A[pos] - A[pos + 1]) + solve(pos + 1));

    if(pos + 2 < n) ans = min(ans, abs(A[pos] - A[pos + 2]) + solve(pos + 2));

    return dp[pos] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    printf("%d\n", solve(0));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3863050608142695x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;



int h[100055],dp[100055];

const int inf = 0x3f3f3f;

int main()

{

	int n;

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

	cin>>n;

	h[1]=inf;

	h[0]=inf;

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

		cin>>h[i];

	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]));

	}

		

	cout<<dp[n]<<endl;

	return 0;

} 

//max(h[i],h[i-2])-min(h[i],h[i-2])

// Faster C++ solution for the same problem (measured speedup: 1.4018937466274897x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

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

typedef long long ll;

const int maxn = 1e6 + 5;

const int inf = 0x3f3f3f3f;

ll a[maxn];

ll dp[maxn];



int main()

{

    IO;

    int n, k=2;

    cin >> n ;

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

        cin >> a[i];

    memset(dp, inf, sizeof(dp));

    dp[1] = 0;

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

    {

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

        {

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

        }

    }

    cout << dp[n] << endl;

}


// Faster C++ solution for the same problem (measured speedup: 9.39555033157816x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

const int nax = 1e5+3;

int number;

int dyn[nax];

int a[nax];

int op1(int i){

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

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

	int res = abs(a[i]-a[i+1])+op1(i+1);

	if(i!=number-2) res = min(res, abs(a[i]-a[i+2])+op1(i+2));

	return dyn[i]=res;

}

int main(){

	cin>>number;

	for(int i=0; i<number; i++) cin>>a[i];

	memset(dyn, -1, sizeof dyn);

	cout<<op1(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3985963364871792x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==1 || x == 0){

		return 0;

	}

	if (x == 2){

		return abs(arr[x] - arr[x-1]);

	}

	if (memo[x] != -1){

		return memo[x];

	}

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

	memo[x] = best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

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

	cin.tie();

	cout.tie();

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.4144654703963049x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

int n, arr[100005], best, memo[100005];



int dp(int x){

	if(x==0 or x==1){

		return 0;

	}

	else if(x==2){

		return dp(x-1)+abs(arr[x]-arr[x-1]);

	}

	else if(memo[x]!=-1){

		return memo[x];

	}

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

	memo[x]=best;

	return best;

}



int main(){

	ios_base::sync_with_stdio(0);

	cin.tie();

	cout.tie();

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

	cin >> n;

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

		cin >> arr[i];

	}

	cout << dp(n) << endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.4144654703963049x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

using namespace std;

int arr[100010],n;

int dp[100010];

int f(int pos){

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

	int &ret=dp[pos];

	if(ret^-1)return ret;

	if(pos+1==n-1)return ret=abs(arr[pos+1]-arr[pos]);

	return ret=min(f(pos+1)+abs(arr[pos]-arr[pos+1]),f(pos+2)+abs(arr[pos]-arr[pos+2]));

}



int main(int argc, char const *argv[])

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

		cout<<f(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.393797164756946x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





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

#define ll long long

using namespace std;

const int maxn = 1e5+5;

int a[maxn], dp[maxn];



int main(){

	int n; cin >> n;

	memset(dp, 0x3f, sizeof dp);

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

	dp[1] = 0;

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

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

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

	}

	cout << dp[n] << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.3985765927051284x):
#include<bits/stdc++.h>

#define ull unsigned long long

#define ll long long

#define pb push_back

#define endl '\n'

#define line cout<<"----------------------"<<endl

#define mp make_pair

#define mod 1000000007

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

#define filein freopen("input.txt","r",stdin)

#define fileout freopen("output.txt","w",stdout)

using namespace std;

const ll  mx=300009;

int ara[mx];

ll  cost[mx];

int  main()

{

    ll  n;

    cin>>n;

    for(ll  i=1;i<=n;i++) cin>>ara[i];

    cost[1]=0;

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

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

    {

        ll  a=cost[i-1]+abs(ara[i]-ara[i-1]);

        ll  b=cost[i-2]+abs(ara[i]-ara[i-2]);

        cost[i]=min(a,b);

    }

    cout<<cost[n]<<endl;





}