c

Farmer John’s cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump.

The local witch doctor has mixed up P (1 <= P <= 150,000) potions to aid the cows in their quest to jump. These potions must be administered exactly in the order they were created, though some may be skipped.

Each potion has a ‘strength’ (1 <= strength <= 500) that enhances the cows’ jumping ability. Taking a potion during an odd time step increases the cows’ jump; taking a potion during an even time step decreases the jump. Before taking any potions the cows’ jumping ability is, of course, 0.

No potion can be taken twice, and once the cow has begun taking potions, one potion must be taken during each time step, starting at time 1. One or more potions may be skipped in each turn.

Determine which potions to take to get the highest jump.
Input

• Line 1: A single integer, P

• Lines 2…P+1: Each line contains a single integer that is the strength of a potion. Line 2 gives the strength of the first potion; line 3 gives the strength of the second potion; and so on.
  Output

• Line 1: A single integer that is the maximum possible jump.
  Sample Input
  8
  7
  2
  1
  8
  4
  3
  5
  6
  Sample Output
  17
  这个题很有意思，有个牛想跳但是他不会跳，所以他要和药水，但是已有奇数的时候可以跳起来，但是偶数的时候不能跳只能跌下来。
  我刚开始想的时候使这样想的，开一个4位的dp，这个dp是存在了四项1.是保存前一个数字是奇数并且取这个数字，2.保存前一个数字不去并且前前个是奇数3.是保存前一个数字是偶数并且取他，4.保存前一个是偶数并且不取他，但是这个是要超时的。
  另一个思路就是开两个dp，一个是存偶数的最大值一个是奇数的最大值这样就简单了。
  dp[i][0]=max(dp[i-1][0],dp[i-1][1]-a[i]);
  dp[i][1]=max(dp[i-1][1],dp[i-1][0]+a[i]);
  上代码

#include<iostream>
#include <cstdio>
#include<algorithm>
#include <cstring>
using namespace std;
int a[160000],dp[160000][2];
int main()
{
    int n;
    while(scanf("%d",&n) != EOF)
    {
        dp[0][0]=dp[0][1]=0;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            dp[i][0]=max(dp[i-1][0],dp[i-1][1]-a[i]);
            dp[i][1]=max(dp[i-1][1],dp[i-1][0]+a[i]);
        }
        cout<<max(dp[n][0],dp[n][1])<<endl;
    }
}