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

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

#define fore(i,a,b) for(int i = a; i < b; ++i)

#define N_MAX 100005

using namespace std;

typedef long long ll;

typedef pair<int,int> pi;

typedef vector<int> vi;

int n,k;

vi v;

vi dp;

void leer(){

    si(n); 

    //si(k);//B

    k=2; //A

    v.assign(n+1,0);

    dp.assign(n+1,INT_MAX);

    fore(i,1,n+1)

        si(v[i]);

    int x;

    x = v[1];

    if(k>n)

        k = n;

    fore(i,1,k+1)

        dp[i] = abs(x-v[i]);

    fore(i,k+1,n+1){

        fore(j,1,k+1){

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

        }

    }

}

void solve(){

	leer();

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

}

int main(){

	solve();

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.310253123681185x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

	for (i=n-3; i>=0; i--) {

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

	}

	

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.335269542524362x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#define f first

#define s second

#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)

#define pb push_back

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

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

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

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

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

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

    }

    else{

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

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.

// Faster C++ solution for the same problem (measured speedup: 4.451302134105037x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

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

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

        cin >> h[i];

    }

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

    // cout << Dp(0) << endl;

    for(int i = n - 2; i >= 0; i--) {

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

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

    }

    cout << dp[0] << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.411571382657487x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#define ll long long

#define pb push_back

#define popb pop_back

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

#define dic unordered_map //Diccionario

#define bpc __builtin_popcountll //numero de bits para long long

#define bclz __builtin_clzll //leading zeros

 

using namespace std;

 

//Funtions here

int main () {

	ios::sync_with_stdio(false);

	cin.tie(0);

	int n;

	cin>>n;

	vector<int> a(n);

	vector<int> dp(n,INT_MAX);

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

	dp[0]=0;

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

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

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

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.423860882826728x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#include<cmath>

using namespace std;

int h[100005];

int dp[100005];

int main()

{

	int n;

	cin>>n;

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

	cin>>h[i];

	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;

}

// Faster C++ solution for the same problem (measured speedup: 4.354524416101873x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#include<algorithm>

#include<cmath>

#include<cstdlib>

#include<cstdio>

#include<cstring>

#include<iomanip>

#include<string>

#include<set>

#include<map>

#include<vector>

#include<queue>

#include<stack>

#include<utility>

#define PI acos(-1.0)

using namespace std;

int main() {

	int n;

	cin >> n;

	vector<int> a(n + 1);

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

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

	a[0] = a[1];

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

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

	}

	cout << dp[n] << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.3633845364736885x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;

const int N=1e5+5;

int v[N],memo[N];

int main(){

	int n;cin>>n;

	if(n==2){

		int x,y;cin>>x>>y;

		cout<<abs(x-y);return 0;

	}

	long long ans=0LL;



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

		cin>>v[i];

		if(i==1)memo[i]=0;	

		else if(i==2)memo[i]=abs(v[i]-v[i-1]);

		

		else{

			memo[i]=min(memo[i-1]+(abs(v[i]-v[i-1])),memo[i-2]+(abs(v[i]-v[i-2])));					

		}

	}

	cout<<memo[n];

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.348432171448303x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;

//-------------------------------------------MACROS---------------------------------------------

#define 	TIEMPO_CORRIDA		cerr << "tiempo: "<<1.0*clock()/CLOCKS_PER_SEC <<" seg.\n";

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

#define 	FORIT(it,e)			for(it = e.begin() ; it != e.end() ; it++)

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

#define 	RFOR(i,a,b)			for(i = a ; i >= b ;i--)

#define 	FORSW(i,a,b,sw)		for(i = a ; i < b && sw ; i++)

#define 	RFORSW(i,a,b,sw)	for(i = a ; i >= b && sw ; i--)

#define 	fast 				ios::sync_with_stdio(false),cin.tie(),cout.tie();

#define 	to_int(x) 			atoi(x.data())

#define 	to_long(x) 			atoll(x.data())

#define 	ALL(x) 				x.begin(),x.end()

#define 	RALL(x) 			x.rbegin(),x.rend()

typedef 	long long 			ll;

typedef 	pair<int,int> 		pi;

typedef 	pair<ll,ll> 		pl;

const		int 				modN = 1e9+7;

const		double 				PI = 3.1415926535897932384626433832795028841;

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

const int tam = 6e5;

const int oo=1e9;

int ar[tam],memo[tam];

int n;



int f(int x){

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

	if(x == n-1) return memo[x]=0;

	int a=oo , b=oo;

	if(x+1 <n) a = f(x+1) + abs(ar[x+1]-ar[x]); 

	if(x+2 <n) b = f(x+2) + abs(ar[x+2]-ar[x]);

	return memo[x]=min(a,b);

}



int main(){

	int i;

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

		memset(memo,-1,sizeof memo);

		FOR(i,0,n) scanf("%d",&ar[i]);

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

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 14.765227455457874x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;

int N;

int main(){

	cin >> N;

	int arr[N];

	int mem[N];

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

		cin >> arr[i];

	mem[0] = 0;

	mem[1] = abs(arr[0] - arr[1]);

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

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

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

}


// Faster C++ solution for the same problem (measured speedup: 4.3583524393625455x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;

 

int main() {

	int n; 

	cin>>n;

	vector<int> arr(n), dp(n);

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

	dp[0]=0;

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

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

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

	}

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

	

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.359474306381025x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

#include <cstdlib>

using namespace std;

int main()

{

    long long n,i;

    static long long ara[100002],t[100002];

    cin>>n;

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

        cin>>ara[i];

    t[0]=0;

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

    {

        if(i==1)

            t[i]=abs(ara[i]-ara[i-1]);

        else

            t[i]=min(t[i-1]+abs(ara[i]-ara[i-1]),t[i-2]+abs(ara[i]-ara[i-2]));

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.349833115801919x):
#include<cmath>

#include<cstdio>

#include<algorithm>



int main(){

  int s;

  long long st[100001],p[100001];

  scanf("%d",&s);

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

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

  }

  p[0]=0;

  p[1]=0;

  p[2]=(st[2]-st[1] < 0) ? st[1]-st[2] : st[2]-st[1];

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

    p[i] = std::min(p[i-1] + ((st[i]-st[i-1] < 0) ? st[i-1]-st[i] : st[i]-st[i-1]) , p[i-2] + ((st[i]-st[i-2] < 0) ? st[i-2]-st[i] : st[i]-st[i-2]));

  }

  printf("%lld\n",p[s]);

  return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

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

  }

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

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.2854848882115817x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

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

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

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

	return memo[n];

}

int main(){

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

	scanf("%d" , &n);

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

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

	}

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

}

// Faster C++ solution for the same problem (measured speedup: 3.6222579077275325x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[100009];

ll n,x;

vector<ll>v;

ll frog(ll pos)

{

    if(pos<=0)

        return 0;

    if(pos==1)

        return abs(v[0]-v[1]);

        if(dp[pos]==-1)

        {

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

        }

    return dp[pos];

}



int main()

{

    cin>>n;

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

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

    {

        cin>>x;

        v.push_back(x);

    }

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

}


// Faster C++ solution for the same problem (measured speedup: 1.2893309904940664x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

int v[1000010];

int memo[1000010];

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: 3.6340762809211813x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



#define F first

#define S second



int N;

int H[100005];

int dp[100005][5];



int solve(int cost, int ind, int j) {

	if (ind > N) return INT32_MAX;

	if (ind == N) return cost;



	if (~dp[ind][j]) return dp[ind][j];

	

	dp[ind][j] = cost;

	dp[ind][j] += min(solve(abs(H[ind] - H[ind + 1]), ind + 1, 0), 

					solve(abs(H[ind] - H[ind + 2]), ind + 2, 1));

	return dp[ind][j];

}



int main() {

	cin >> N; 

	H[0] = INT32_MAX; // to avoid RTE

	for (int i = 1; i <= N; i++) cin >> H[i];

	int ans = 0;

	

	memset(dp, -1, sizeof dp);

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



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.1465878519867103x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>

#define inf INFINITY

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

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

#define _3102_ main()

#define endl '\n'

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

using namespace std;

const int N = 1e5 + 100;

const ll MX = 1e9;

const ll mod = 1e9 + 7;

const double eps = 1e-9 ;



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

ll solve( int i ){

	if ( i < 0 ) return 0 ;

	ll &ret = dp[i] ;

	if ( ~ret ) return ret ;

	if ( i == 0 ) return ret = 0 ;

	if ( i == 1 ) return ret = abs( h[1] - h[0]) ; 

	return ret = min ( solve(i - 1 ) + abs(h[i] - h[i-1]) , solve ( i - 2 ) + abs ( h[i] - h[i-2])) ;

}



int _3102_{

	Gamed

	cin >> n ;

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

		cin >> h[i] ;

	}

	memset ( dp , -1 , sizeof dp ) ;

	cout << solve ( n - 1 ) ;



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.2962392041825057x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH
// Slower C++ solution:
#include <bits/stdc++.h>



#define fst first

#define snd second

#define fore(i,a,b) for(int i=a, ThxDem=b; i<ThxDem; ++i)

#define pb push_back

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

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

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

#define DBG(x) cerr <<#x << " = " << (x) << "\n"



#define SQR(x) ((x) * (x))



#define PI 3.1415926535897932385

#define INF 1000111222

#define eps 1e-7

#define maxN 11



using namespace std;



typedef long long ll;

typedef vector<ll> vll;

typedef pair<ll,ll> pll;

typedef pair<int,int> pii;

typedef vector<string> vstr;

typedef vector<vll> vvll;

typedef vector<bool> vbool;

typedef vector<pll> vpll;

typedef vector<vpll> vvpll;

typedef vector<vbool> vvbool;



vll ns,dp;

ll n;



ll solve(ll pos){

	ll & d = dp[pos];

	if(d != -1) return d;

	ll v=INF;

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

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

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

	return d = v;

}



int main(){FIN

	

	cin >> n;

	ns.resize(n);

	dp.assign(100010,-1);

	fore(i,0,n)cin >> ns[i];

	

	solve(0);

	

	cout << dp[0] << "\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.304164635598037x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

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

int main()

{

	cin>>n;

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

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

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

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH