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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long
  5. const int MOD = 158400;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX/2;
  8.  
  9. int primes[1000000];
  10.  
  11. void seive(){
  12. fill(primes, primes + 1000000, 1);
  13. primes[0] = primes[1] = 0;
  14. for(int i = 2 ; i*i < 1000000 ; i++){
  15. if(primes[i]){
  16. for(int j = i*i ; j < 1000000 ; j += i){
  17. primes[j] = 0;
  18. }
  19. }
  20. }
  21. for(int i = 1 ; i < 1000000 ; i++){
  22. primes[i] += primes[i-1];
  23. }
  24. }
  25. int factorial(int n){
  26. if(n==0){
  27. return 1;
  28. }
  29. return (n*(factorial(n-1)))%MOD;
  30. }
  31. bool isPrime(int n){
  32. if(n <= 1) return false;
  33. for(int i = 2 ; i*i <= n ; i++){
  34. if(n % i == 0) return false;
  35. }
  36. return true;
  37. }
  38.  
  39. int power(int a, int b){
  40. if(b == 0) return 1;
  41. a %= MOD;
  42. int value = power(a, b / 2);
  43. if(b % 2 == 0){
  44. return (value * value) % MOD;
  45. } else {
  46. return ((value * value) % MOD * (a % MOD)) % MOD;
  47. }
  48. }
  49.  
  50. int gcd(int a, int b){
  51. if(a == 0) return b;
  52. return gcd(b % a, a);
  53. }
  54. void solve() {
  55. int n;
  56. cin>>n;
  57. int A[n];
  58. for(int i = 0 ; i<n ; i++){
  59. cin>>A[i];
  60. }
  61. int minimumCost[n];
  62. minimumCost[0] = 0;
  63. minimumCost[1] = abs(A[1]-A[0]);
  64. for(int i = 2 ; i<n ; i++){
  65. int d = abs(A[i]-A[i-1])+minimumCost[i-1];
  66. int e = abs(A[i]-A[i-2])+minimumCost[i-2];
  67. minimumCost[i] = min(d,e);
  68. }
  69. cout<<minimumCost[n-1]<<endl;
  70. }
  71.  
  72. signed main(){
  73. ios::sync_with_stdio(false); cin.tie(NULL);
  74. //int t;
  75. //cin >> t;
  76. //while(t--){
  77. solve();
  78. //}
  79. return 0;
  80. }
  81.  
  82.  
Success #stdin #stdout 0.01s 5292KB
stdin
6
30 10 60 10 60 50
stdout
40