POJ-2229

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 19599		Accepted: 7651

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

题意：
给出一个整数n,求n有多少种由2的幂次之和组成的方案. 

当n为奇数的时候，那么所求的和式中必有1，则dp[n]==dp[n-1]；

当n为偶数的时候，可以分两种情况：

1.含有1，个数==dp[n-1]；

2.不含有1，这时每个分解因子都是偶数，将所有分解因子都除以二，所得的结果刚好是n/2的分解结果，并且一一对应，则个数为dp[n/2]；

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
using namespace std;

const long long MOD=1000000000;

int dp[1000010];

int main(){
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n&&n){
        memset(dp,0,sizeof(dp));
        dp[1]=1;
        for(int i=2;i<=n;i++){
            if(i&1){
                dp[i]=dp[i-1];
            }
            else{
                dp[i]=(dp[i-1]+dp[i>>1])%MOD;
            }
        }
        cout<<dp[n]<<endl;
    }
    return 0;
}