poj 2220 Sumsets

                                                                                                 Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 16876		Accepted: 6678

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

初学动态规划，我用了一种非常愚蠢的解法耗内存又超时了...
AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
int dp[1000000+1];
/*const int N_MAX = 1000000;
int dp[21][N_MAX + 1];
int main() {
    int N;
    while (cin >> N) {
        int k = 0;
      while ((1 << k) <= N) {//求使得2^k大于N的最小k
        k++;
     }

      for (int i = 0;i < k;i++) 
          dp[i][0] = 0;
      for (int i = 1;i <= N;i++)
          dp[0][i] = 1;
      for (int i = 1;i < k;i++) {
          for (int j = 1;j <= N;j++) {
              dp[i][j] = dp[i - 1][j];
              for (int k1 = 1;(j - k1*(1 << i))>=0;k1++) {
                  dp[i][j] += dp[i - 1][j - k1*(1 << i)];
              }
          }
      }
      cout << dp[k-1][N] << endl;
    }
    return 0;
}*/
//若i为奇数,(i-1)为偶数，i的组合数就是(i-1)的组合数，因为(i-1)只能加1得到i。若i为偶数，(i-1)为奇数，则通过(i-1)+1的方式得到i的组合必定带有1，接下来考虑
//全是偶数的组合数，考虑到全是偶数的组合数和(i/2)的组合数一样，因为只要(i/2)的组合数里每一个数*2就可以得到i
int main() {
    int N;
    while (cin >> N) {
        dp[1] = 1;
        for (int i = 2;i <= N;i++) {
            if ((i & 1)==0) {//若为偶数
                dp[i] = dp[i / 2];
            }
            dp[i] += dp[i - 1];
            dp[i] %= 1000000000;
        }
        cout << dp[N] << endl;
    }
    return 0;
}