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=;

 int dp[];

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