多项式的猜想

#include <stdio.h>#include <stdlib.h>int getA ( int n ){                     //得到每一项    if ( n == 1 || n == 2 )    {        return 1;    }    else    {        return 2 * getA ( n - 1 ) + getA ( n - 2 );    }}int getS ( int n ){             //得到前n项的和    int i, s = 0;    for ( i = 1; i <= n; i++ )    {        s += getA ( i );    }    return s;}int main(){    int j, k = 100;    while ( k <= 10000 )    {        for ( j = 1; j < 10000; j++ )        {            if ( getS ( j ) < k && getS ( j + 1 ) >= k )            {                printf ( "%d\n", j );                break;            }        }        k *= 10;    }    return 0;}