URAL 1079. Maximum

problem url:  http://acm.timus.ru/problem.aspx?space=1&num=1079

this should be a mathematics question. So far I got two answers. MY own is using static array to cache all the sequence elements while there is another mathematic way which dosn't require Array or cache(of each sequence element):

#simple approach(not good at scalability):

#include  < cstdio >
#include  < iostream >


#define  MAXNUM 100000

using   namespace  std;

// GLOBAL VARIABLES
int  NUMS[MAXNUM / 2 + 2 ];  // each item of the sequence

int  curmax  =   0 ;

int  inputs[ 10 ];
int  maxs[ 10 ];
int  inputcnt  =   0 ;
int  maxinput  =   0 ;

void  main()
... {
    int i =0,j=0;
    int curinput=0;    

    do
    ...{
        scanf("%d", &curinput);
        inputs[inputcnt++] = curinput;

        if(curinput > maxinput) maxinput = curinput;

    }while(curinput != 0);

    curmax =0;
    int curitem = 0;

    for(i=0;i<= maxinput;++i)
    ...{
        if(i==0) curitem = 0;
        else if(i == 1) curitem = 1;
        else
        ...{
            if(i%2)
            ...{
                curitem = NUMS[i/2] + NUMS[i/2 +1];
            }else
            ...{
                curitem = NUMS[i/2];
            }
        }

        if(curmax < curitem) curmax = curitem;

        if(i<(MAXNUM/2+2)) NUMS[i] = curitem;

        for(j=0;j<inputcnt-1;++j)
        ...{
            if(inputs[j] == i) maxs[j] = curmax;
        }
    
    }
    
    for(i=0;i<inputcnt-1;++i)  printf("%d ", maxs[i]);
}

#pure mathematics approach(can not understand yet :( ):

/*

Assume g(n,i,j)=i*f(n)+j*f(n+1).
Then
g(2n,i,j)=g(n,i+j,j)
g(2n+1,i,j)=g(n,i,i+j)
We must find f(n)=g(n,1,0), so... 

*/


#include  < iostream >
int     n,x;
int     a[ 10 ];m
int     i,t;

int     _max( int  s1, int     s2, int  x)
{
     if  (x == n)
         return  s1 + s2;
     else  {
         int     t1,t2;
         if  (x * 2 - 1 <= n)
            t1  =  _max(s1,s2 + s1,x * 2 - 1 );
         else
            t1  =   0 ;
         if  (x * 2 + 1 <= n)
            t2  =  _max(s1 + s2,s2,x * 2 + 1 );
         else
            t2  =   0 ;
         if  ((t1 == t2) && (t2 == 0 ))
             return  s1 + s2;
         else
             return  t1 > t2 ? t1:t2;
    }
}

int     main()
{
    std::cin  >>     n;
    i  =      0 ;
     while  (n){
         if  (n == 2 )
            a[i]  =   1 ;
         else  
             if  (n == 1 )
                a[i]  =   1 ;
             else
                 if  (n == 0 )
                    a[i]  =   0 ;
                 else
                    a[i]  =  _max( 1 , 1 , 3 );
        std::cin  >>     n;
        i ++ ;
    }
     for     (n  =   0 ;n < i;n ++ )
        std::cout  <<  a[n]  <<   " " ;

     return   0 ;
}