/*51nod 1270 数组的最大代价problem:数组A包含N个元素A1, A2......AN。数组B包含N个元素B1, B2......BN。并且数组A中的每一个元素Ai，都满足1 <= Ai <= Bi。数组A的代价定义如下：两个相邻元素A的差的绝对值之和solve:按照贪心的思路来,感觉肯定是取 B[i]或者1.  所以第i步的情况应该是由i-1推过来的.但是写的时候,觉得如果第i位如果取最大值,那么i-1就取最小值mdzz,所以少了一些情况. 实际上i-1的最大最小值都应该考虑.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) scanf("%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;}