Maximum cost of 51nod 1270 arrays

1270 the maximum cost of an array
Title Source: Hackerrank
Base time limit: 1 seconds space limit: 131072 KB score: 20 Difficulty: 3-level algorithm problemarray A consists of n elements A1, A2 ... An. Array B consists of n elements B1, B2 ... BN. And each element of the array, AI, satisfies 1 <= ai <= Bi. The cost of array A is defined as follows:  (the formula represents the sum of the absolute values of the difference of all two adjacent elements)give the array b to calculate the possible maximum cost S. Input

Line 1th: 1 number N, indicating the length of the array (1 <= n <= 50000). Line 2-n+1:1 numbers per line, corresponding to array element bi (1 <= bi <= 10000).

Output

The maximum cost of output s.

Input Example

510110110

Output Example

36

/*51nod the maximum cost of an array of 1270 problem: array A contains n elements A1, A2 ... An. Array B consists of n elements B1, B2 ... BN. And each element of the array, AI, satisfies 1 <= ai <= Bi.  The cost of array A is defined as follows: The sum of the absolute values of the difference of two adjacent elements a solve: according to the greedy idea, the feeling must be take b[i] or 1. So the situation of step I should be pushed by i-1. But when writing, I think if the first bit if the maximum value, then i-1 take the minimum value mdzz, so there are a few cases. In fact, the maximum minimum value of i-1 should be considered. hhh-2016/09/03 13:32:28*/#pragma COMMENT (linker, "/stack:124000000,124000000") #include < algorithm> #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring># Include <vector> #include <math.h> #include <queue> #include <set> #include <map> #define Lson i<<1#define Rson i<<1|1#define ll long Long#define CLR (A, B) memset (A,b,sizeof (a)) #define SCANFI (a) scan F ("%d", &a) #define SCANFS (a) scanf ("%s", a) #define SCANFL (a) scanf ("%i64d", &a) #define SCANFD (a) scanf ("%lf",  &a) #define KEY_VAL ch[ch[root][1]][0] #define EPS 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1e9+7;const int maxn = 50010;const Double PI = ACOs ( -1.0); ll dp[maxn][2];ll A[maxn];int Main () {int n;        while (SCANFI (n)! = EOF) {clr (dp,0);        for (int i = 1;i <= n;i++) {SCANFL (a[i]);        }//dp[1][1] = a[1];//dp[1][0] = 1;            for (int i =2; i<=n;i++) {dp[i][1] = Dp[i-1][0] + a[i]-1LL;            Dp[i][0] = dp[i-1][1] + a[i-1]-1LL;            DP[I][1] = max (dp[i-1][1] + (A[i]-a[i-1]), dp[i][1]);        Dp[i][0] = max (dp[i-1][0],dp[i][0]);    } printf ("%i64d\n", Max (dp[n][0],dp[n][1])); } return 0;}

　　

Maximum cost of 51nod 1270 arrays