stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 5.282395470621915x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// 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: 5.303975018796527x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// 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: 5.205561545677105x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// 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: 5.2416727225017405x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// Slower C++ solution:
#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;

} 

// Faster C++ solution for the same problem (measured speedup: 3.0910827113425294x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 9.364534690829958x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 5.231011472822243x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<algorithm>

using namespace std;

int main(){

    int dp[100001]={0};

    int stone[100001];

    int n;

    scanf("%d",&n);

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

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

    }

    dp[1] = 0;

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

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

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

    }

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

}


// Faster C++ solution for the same problem (measured speedup: 3.1875838857732246x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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



using namespace std;

int v[100010];

int memo[100010];

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()

{

    ios_base::sync_with_stdio(false);

    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: 7.411673398191387x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

using namespace std;

#define Me3za() ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);srand(time(NULL))

#define endl "\n"

int n;

int arr[100005];

int mem[100005];

long long solve(int i,int cur){

if(i>=n)return 1e9;

//cout<<i<<" "<<cur<<endl;

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

long long ch1=solve(i+1,arr[i+1])+abs(cur-arr[i+1]);

long long ch2=0;

if(i+2<=n)ch2=solve(i+2,arr[i+2])+abs(cur-arr[i+2]);

return mem[i]=min(ch1,ch2);

}

int main() {

Me3za();

cin>>n;

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

    cin>>arr[i];

}

memset(mem,-1,sizeof mem);

cout<<solve(0,arr[0]);

}




// Faster C++ solution for the same problem (measured speedup: 7.386823340639693x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

using namespace std;



int const N=1e5+5, M=1e9+7, OO = 0x3f3f3f3f;



int n;

int arr[N];



int mem[N];

int solve(int i)

{

    if(i==n-1)

        return 0;

    if(mem[i]!=-1)

        return mem[i];

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=(i<=n-3) ? solve(i+2)+abs(arr[i]-arr[i+2]) : OO;

    return mem[i]=min(op1,op2);

}





int main()

{

//    freopen("input.txt","rt",stdin);

//    freopen("output.txt","wt",stdout);

    scanf("%d",&n);

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

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

    memset(mem, -1, sizeof mem);

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.262048862777216x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}



const int N = 1e5 + 9;

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



int sol(int i){

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

    if(i == n)  return 1e9;



    int &ret = dp[i];

    if(~ret)    return ret;



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

}





int main() {



    cin >> n;



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



    memset(dp, -1, sizeof dp);



    cout << sol(0);

}

// Faster C++ solution for the same problem (measured speedup: 7.317636881516436x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<algorithm>

using namespace std;

const int maxn=100001;

int a[maxn],dp[maxn];



int main(){

	int n;

	cin>>n;

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

		cin>>a[i];

	}

	dp[n]=0;

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

	for(int i=n-2;i>=1;i--){

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

	}

	cout<<dp[1];

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.224159330401822x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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



using namespace std;



const int N=1e5;

int n;

int arr[N];

int mem[N];



int solve(int i){

    if(i==n-1)

        return 0;

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

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=i+2<=n-1?solve(i+2)+abs(arr[i]-arr[i+2]):1000000000;

    return mem[i]=min(op1,op2);

}



int main() {

    scanf("%d",&n);

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

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

    memset(mem,-1,sizeof mem);

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 7.260261117572121x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<iostream>

#include<iomanip>

#include<map>

#include<cmath>

#include<algorithm>

#include<string>

#include<vector>

#include<queue>

#include<set>

#include<deque>

#include<stack>

#include<bitset>

#include<fstream>

#include<map>

#include<unordered_map>

using namespace std;

typedef int me;

typedef long long ll;

typedef unsigned long long ull;

#define andrew  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

#define endl '\n'

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

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

#define PI 3.1415926535

#define sin(i) (long double)sin(i*PI/180.0);

ll gcd(ll x, ll y) { return (y == 0) ? x : gcd(y, x % y); }

ull lcm(ll x, ll y) { return x / gcd(x, y) * y; }

void the_end() {

#ifndef ONLINE_JUDGE;

    system("pause");

#endif

}

int dy8[] = { 0,0,1,-1,1,-1,-1,1 };

int dx8[] = { 1,-1,0,0,1,-1,1,-1 };

int id8[] = { 0, 1,2,3,4, 5,6, 7 };

// you can be specialist 

// soon will be (Cyain)

ll n;

vector<ll>v;

map<int, ll>m;

ll solve(int i)

{

    if (i == 0)

        return 0;

    if (i == 1) {

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

    }

    if (m.find(i) != m.end())

        return m[i];

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

}

int main()

{

    andrew;

    cin >> n;

    v.resize(n);

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

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

    the_end();

}

// Faster C++ solution for the same problem (measured speedup: 5.369235154601572x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

	cin.tie(NULL), cout.tie(NULL);

}





int main()

{

	fast();



	int n;

	cin >> n;

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

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

		cin >> v[i];

	}

	dp[1] = 0;

	dp[2] = dp[1] + abs(v[1] - v[2]);

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

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

	}

	cout << dp[n];





	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.304598053285698x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



int n,a[100050];long long mem[100050];



long long solve(int idx)

{

    if(idx==n-1)

        return 0;

    if(idx>=n)

        return 1500000000;

    if(mem[idx]!=-1)

        return mem[idx];

    int op1,op2;

    //if(idx<n-1)

        op1=solve(idx+1)+abs(a[idx]-a[idx+1]);

    //if(idx<n-2)

        op2=solve(idx+2)+abs(a[idx]-a[idx+2]);

    return mem[idx]=min(abs(op1),abs(op2));

}



int main() {



    cin>>n;

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

    {

        cin>>a[i];

        mem[i]=-1;

    }

    solve(0);

    cout<<mem[0];



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.2234193328681515x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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

#include<utility>

#include<iomanip>

#include<cmath>

#include<algorithm>

#include<cfloat>

#include<functional>

#include<string>

#include<numeric>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#include<cctype>

#include<iterator>

#include <stdio.h>

#include <string.h>

#include <sstream>

// #include <bits/stdc++.h>

//#include<conio.h>

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

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

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

#define M_PI 3.14159265358979323846

#define mod 1000000007

#define endl "\n"

using namespace std;

using ll = long long;

using lli = long long int;

using ld = long double;





const int N = 5e5 + 6, inf = 2e9;

int n;

int arr[N], dp[N];

int main()

{

	MEGA_MIND



	cin >> 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(abs(arr[i - 1] - arr[i]) + dp[i - 1], abs(arr[i - 2] - arr[i]) + dp[i - 2]);

	}

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.304813003276863x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

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



using namespace std;



int main()

{

    int n;

    cin>>n;

    int arr[n+5],dp[n+5];

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

        cin>>arr[i];

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

        dp[i]=1e9;



    dp[0]=0;

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

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

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

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.231155578714829x):
#include <stdio.h>

#include <math.h>

#include <stdlib.h>



int min(int a, int b)

{

    return a < b ? a : b;

}

int main()

{

    int n;

    scanf("%d", &n);

    int a[n];

    int dp[n];

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

    {

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

    }

    dp[0] = 0;

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

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

    {

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

    }

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

}