POJ 2181 Mm... train of thought questions

Previously, the sum of the largest continuous subsequences was developed by myself. linear algorithms are quite simple. I haven't updated my problem solution for a long time. I don't know when I will be able to come out of the solution for 1,000, so I have a good question. In a sequence, take the numbers. The odd digits are the plus digits, and the even digits are the minus digits. The greatest sum is required. I had no idea, and I was not planning to think about it. I was too tired at noon. I had A rest and thought about it after I got up. I wrote 1A and wrote A again. Well, this is A question of water. I only want to cheat =

#include<iostream>#define MAXN 150001using namespace std;int date[MAXN];int main(){    int n,i;    while( scanf( "%d",&n )!=EOF )    {           date[0]=0;           for( i=1;i<=n;i++ )                scanf( "%d",&date[i] );           int ans=0;           i=0;           while( date[i+1]>=date[i] )                  i++;           ans+=date[i];                      for(;i<=n;i++ )           {                while( date[i+1]<=date[i] && i<=n ) i++;                if( i!=n )                      ans-=date[i];                 while( date[i+1]>=date[i] && i<=n ) i++;                       ans+=date[i];           }           printf( "%d\n",ans );               }    return 0;}

#include<iostream>#define MAXN 150001using namespace std;int max( int a,int b ){ return a>b?a:b; }int main(){    int n,i;    while( scanf( "%d",&n )!=EOF )    {           bool add=true;           int a,b;           a=0;int ans=0;           scanf( "%d",&a );           for( i=2;i<=n;i++ )           {                scanf( "%d",&b );                if( add )                {                    if( a>b )                    {                             add=!add;                         ans+=a;                    }                    if( b>=a && i==n )                        ans+=b;                }                else                {                    if( a<b )                    {                             add=!add;                         ans-=a;                    }                    if( a<b && i==n )                        ans+=b;                }                a=b;           }           printf( "%d\n",ans );               }    return 0;}

But it's still not fast to write it out... it's fast !!