挑战程序设计竞赛 2.3章习题 poj 2229 Sumsets dp

// 112355551111.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

#include <iostream>


using namespace std;


/*
将一个数N分解为2的幂之和共有几种分法？
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).
*/


const int N = 1000010;
int dp[N];
int n;

int main() {
    cin >> n;
    dp[0] = 1; dp[1] = 1; dp[2] = 2;
    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1];
        if ((i & 1 ) == 0) {
            dp[i] += dp[i / 2];
        }
        dp[i] = dp[i] % 1000000000;
    }
    cout << dp[n] % 1000000000 << endl;

    return 0;
}