POJ 2229 Sumsets(计数dp)

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

分析

dp[i]:数字i的最多组合数
若i是奇数：i的任意一个组合都包含1，所以dp[i] = dp[i-1]
若i是偶数：i-1是奇数，任意一个组合加1构成i的组合，除了这些组合之外，i的其他组合都不包含1，意味着全部是偶数，则i/2的每个组合乘以2可以构成i的一个组合
所以dp[i]=dp[i-1]+dp[i>>1]

/*===============================================================*   Copyright (C) 2016 All rights reserved.*   *   文件名称：B.cpp*   创 建 者：gsh*   创建日期：2016年08月10日*   描    述：2的幂数 dp*   更新日志：*================================================================*/#include <iostream>#include <cstdio>typedef long long ll;typedef unsigned long long ull;using namespace std;const int maxn = 1000001;const int MOD = 1000000000;ll a[maxn];int main(){    int n;    while(scanf("%d",&n) != EOF)    {         a[0] = 1;         for(int i = 1; i <= n; i++)         {            if(i&1) a[i] = a[i-1];            else a[i] = (a[i-1] + a[i >> 1]) % MOD;         }        printf("%d",a[n]);    }    return 0;}