stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.8176497588735985x):
#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;

}
// 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: 1.8287631848947588x):
#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;

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

using namespace std;



#define fi first

#define se second

#define pb push_back

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

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



typedef long long ll;

typedef long double ld;

typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

typedef vector<int> vi;

typedef vector<vi> vvi;

typedef vector<pii> vii;



const int MAX = 1e5;

const int INF = 1e9;

vi stone(MAX + 5);

vi dp(MAX + 5, -1);

int n;



int sol(int i){

    if(i >= n) return INF;

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

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

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

    if(i + 2 < n){

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

    }

    return dp[i];

}



int main(){

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    #ifdef LOCAL

        freopen("input.txt", "r", stdin);

    #else

        #define endl '\n'

    #endif

    cin >> n;

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

        cin >> stone[i];

    cout << sol(0) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8356678435319078x):
#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;

}
// 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: 3.3944353869161743x):
#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;

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

using ll = long long;

using namespace std;



int n;

ll dp[100100];

int h[100100];

bool vis[100100];



ll func( int pos ) {

    if (pos == n) return 0;

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

    vis[pos] = 1;



    ll p = 2e9, q = 2e9;

    if ( pos + 1 <= n ) p = abs(h[pos] - h[pos + 1]) + func(pos + 1);

    if ( pos + 2 <= n ) q = abs(h[pos] - h[pos + 2]) + func(pos + 2);



    return dp[pos] = min(p, q);

}





int main() {

    cin >> n;

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

    cout << func(1) << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.352667110337071x):
#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 =1e6;



int dp[N], h[N];



int main(){

    int n;

    cin >> n;

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

        cin >> h[i];

    }

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

    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-1] << endl;



}


// Faster C++ solution for the same problem (measured speedup: 4.355482676782218x):
#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, i;

    scanf("%d",&n);

    int height[n+5];

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

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

    }

    int ans = 1e9;

    int dp[n+5];

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

        dp[i] = 1e9;

    }

    dp[1] = 0;

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

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

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

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.302480635031747x):
#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 MAXN = 1e5 + 1;



int n, k, h[MAXN];



int main(){



    cin >> n;

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

    int dp[n + 5];

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

    dp[1] = 0;

    for(int i = 2; i <= n; i++) for(int j = 1; j <= 2 && j < i; j++) if(i > j) dp[i] = min(dp[i], dp[i - j] + abs(h[i] - h[i - j]));

    cout << dp[n] << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.355501409081964x):
#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>



using namespace std;

typedef long long int ll;

ll dp[100010];

ll arr[100010];



int main(){

    ll n;

    scanf("%lld",&n);



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

        scanf("%lld",arr+i);

    }



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

        dp[i] = 10000000000000;

        if(i-1 >= 1){

            dp[i] = dp[i-1] + llabs(arr[i]-arr[i-1]);

        }

        if(i-2 >= 1){

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

        }

    }

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

}


// Faster C++ solution for the same problem (measured speedup: 4.299360332329952x):
#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 arr[100010];

int dp[100010];



int main (){

	int n;

	cin >> n;

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

		cin >> arr[i];

	}

	dp[1] = 0;

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

	

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

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

	}

	

	cout << dp[n] << endl;

}

// Faster C++ solution for the same problem (measured speedup: 4.35437132840051x):
#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 <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

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

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;



ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 - ch + '0';

        }

    }

    else {

        *p = ch - '0';

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 + ch - '0';

        }

    }

}

// Faster C++ solution for the same problem (measured speedup: 4.4055400852552475x):
#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 mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.099605113527038x):
#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: 7.8408766486953985x):
#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<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

using namespace std;

int a[100005],d[100005];

int main()

{

	int n,i;

	memset(a,0,sizeof(a));

	scanf("%d",&n);

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

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

	d[n] = 0;

	d[n - 1] = abs(a[n] - a[n - 1]);

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

	{

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

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.5888957898998366x):
#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 , 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: 22.094135609611573x):
#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(){

	long long n,a[100000],dp[100000];

	scanf("%lld", &n);

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

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

	}

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

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

	for(long long 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", dp[n-1]);

}


// Faster C++ solution for the same problem (measured speedup: 4.464282377589599x):
#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 X first

#define Y second

#define MP make_pair

#define ld long double

 

using namespace std;

 

const int N = 5e5 + 123;

const ll mod = 1e9 + 7;



ll a[N], dp[N];

int n;



int main (){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);



	cin >> n;

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

		cin >> a[i];

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

		int x = i - 1, y = max(1, i - 2);

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

	}

	cout << dp[n];

}

// Faster C++ solution for the same problem (measured speedup: 4.417243712566772x):
#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 int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

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

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

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

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

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 2e5 + 48, mod = 1e9 + 7;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





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



main () {

	cin >> n;

	forn (i, 1, n)

		cin >> h[i];	

	forn (i, 2, n)

		dp[i] = min(dp[i - 1] + abs(h[i] - h[i - 1]), i < 3 ? inf : dp[i - 2] + abs(h[i] - h[i - 2]));		



	cout << dp[n];

} 

// Faster C++ solution for the same problem (measured speedup: 4.348360307747465x):
#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;

}