【POJ】Sumsets

Problem here

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

Solution

當n為奇數時，dp[n] = dp[n-1] 
當n為偶數且含有1時，dp[n] = dp[n-1] 
當n為偶數且不含有1時，dp[n]=dp[n/2] 
所以當n為偶數時，dp[i] = (dp[i-1] + dp[i/2])%1000000000

#include <iostream>
#include <memory.h>
using namespace std;

int dp[1000001];

int main(){
    dp[0] = 0;
    dp[1] = 1;
    int n;
    while(cin >> n){
        for(int i = 2; i <= n; i++){
            if(i % 2 == 0)
                dp[i] = (dp[i-1] + dp[i/2])%1000000000;
            else
                dp[i] = dp[i-1];
        }
        cout << dp[n] << endl;
    }

    return 0;
}