2229:Sumsets

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 21492		Accepted: 8355

Description

Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

1) 1+1+1+1+1+1+1 
2) 1+1+1+1+1+2 
3) 1+1+1+2+2 
4) 1+1+1+4 
5) 1+2+2+2 
6) 1+2+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 

Input

A single line with a single integer, N.

Output

The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).

Sample Input

7

Sample Output

6

Source

USACO 2005 January Silver

递推式：

若n为奇数，则f(n) = f(n-1);

若n为偶数，则n的所有等式中，有一部分全部由偶数构成，另一部分则含有1，且1的数目为偶数个，对于前者，其数量等于f(n/2)，对于后者，其数量等于f(n-1)，则f(n) = f(n-1) + f(n/2)。 

#include<cstdio>
using namespace std;
const int maxn = 1000005,nn = 1000000000;
int n,dp[maxn];
void solve(){
    scanf("%d",&n);
    dp[1] = 1;
    for(int i = 2;i <= n;i++){
        if(i % 2) dp[i] = dp[i-1] % nn;
        else dp[i] = dp[i-1] + dp[i/2] % nn;
    }
    printf("%d\n",dp[n]);
}
int main()
{
    solve();
    return 0;
}