poj 2229

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POJ 2229

#include <iostream>
#include <cmath>
using namespace std;
/*
if n is odd ,  then f(n) = f(n-1) .because n is odd , add the answer must have at least a 1, and f(n-1) above all solutions
else if n is even , then f(n) = f(n-2) + f(n/2) . because n is even ,and must have at least two 1 , then f(n-2)+1+1 above all answers that include 1.
and we must calculate the number of solution (like 2,2,2,2 ==8; 2,2,4==8;4,4=8)
 without 1 . then we can div 2 ,for  2/2 -> 1 .  change(1,1,1,1 ; 1,1,2;2,2 ; 4familiar == 4)
so we can get the answer  f(n) = f(n-2) + f(n/2) .
*/
int mod = 1e9 ;
const int maxn = 1e6+10;
int a[maxn] ;
int main(){
    int t ;
    a[1] = 1 ;
    a[2] = 2 ;
    for(int i = 3 ; i < maxn-9 ; ++i){
        if(i&1) a[i] = a[i-1] ;
        else a[i] = (a[i-2] + a[i>>1])%mod ;
    }
    while(cin >> t){
        cout << a[t] << endl ;
    }
    return 0;
}