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 ;

 }

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POJ-2229

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