POJ 2229 Sunsets

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 11009		Accepted: 4433

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

 

说实话我觉得这道题不应该算动态规划，难道分类到动规的原因是递推公式和记忆化，那好吧。。。

设数字n的表示形式有f(n)种，递推公式就是：

f(n)=f(n-1)  (n为奇数)

f(n)=f(n/2)+f(n-1)  (n为偶数)

因为只能用2^i表示，所以奇数必然要有1，1后面的表示方法数就是f(n-1)了，所以n为奇数时就是f(n)=f(n-1)

而偶数时要分类计论，表示中含有1的和不含有1的，含有1的就是f(n-1)，不含有1的就是f(n/2)，所以n为偶数时就是上f(n)=f(n-1)+f(n/2)

 

#include<iostream>
#include<cstdio>

using namespace std;

const int mod=1000000000;

long long dp[1000010];

int main()
{
    int n;

    dp[1]=1;
    for(int i=2;i<=1000000;i++)
        if(i%2)
            dp[i]=dp[i-1];
        else
            dp[i]=(dp[i-1]+dp[i/2])%mod;

    while(scanf("%d",&n)==1)
        cout<<dp[n]<<endl;

    return 0;
}