ZZULI_TEAM_PRACTICE(1)  POJ 2181…

JumpingCows _p

Time Limit: 1000MSMemory Limit: 65536KB64bit IO Format: %I64d & %I64u

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Description

Farmer John's cowswould like to jump over the moon, just like the cows in theirfavorite 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 questto jump. These potions must be administered exactly in the orderthey 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. Beforetaking any potions the cows' jumping ability is, of course,0.

No potion can be taken twice, and once the cow has begun takingpotions, one potion must be taken during each time step, startingat 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 singleinteger, P

* Lines 2..P+1: Each line contains a single integer that is thestrength of a potion. Line 2 gives the strength of the firstpotion; line 3 gives the strength of the second potion; and soon.

Output

* Line 1: A singleinteger that is the maximum possible jump.

Sample Input

8 

7 

2 

1 

8 

4 

3 

5 

6

Sample Output

17 

这题就没看，不是我做的，不过好像是一道DP题

代码：

C语言: 临时自用代码

#include<stdio.h>
int p[150005],in[150005],de[150005];
int max(int a,int b)
{
    if(a>b)
        return a;
    return b;
}
int main()
{
    int i,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        scanf("%d",&p[i]);
    in[1]=p[1];
    de[1]=0;
    for(i=2;i<=n;i++)
    {
        in[i]=max(de[i-1]+p[i],in[i-1]);
        de[i]=max(in[i-1]-p[i],de[i-1]);

    }
    printf("%d\n",max(in[n],de[i]));
    return 0;
}