51nod 1383&1048 integer decomposition to power of 2 [recursive] "math"

Topic Connection: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1048

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Integer decomposition to a power of 2

Base time limit: 3 seconds Space limit: 131072 KB score: 1280 Difficulty: 9-level algorithm problem

Any positive integer can be decomposed into a power of 2, given an integer n, to find the number of such partitioning methods of N.
For example, n = 7 o'clock, there are 6 ways of partitioning.

7=1+1+1+1+1+1+1
=1+1+1+1+1+2
=1+1+1+2+2
=1+2+2+2
=1+1+1+4
=1+2+4

Input
Input n (1 <= n <= 10^30)

Output
Number of partitioning methods

Input example

7

Output example

6
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First you can DP

Set DP[I][J] indicates that the number of components is I I, the largest number in the scheme is the scheme number of 2j 2^j.
To avoid duplication, Dp[i][j] The next number of enumerations must be greater than or equal to 2j when transferring 2^j
Then you can conclude: Dp[i][j]=∑jk=0dp[i−2j][k] dp[i][j]=\sum^{j}_{k=0}dp[i−2^j][k]

Ability to achieve O (Nlog (n)) O (N\log (n)) complexity

By playing the table to find the law, etc., you can recursively get the following rules

Dp[i]=dp[i−2]+dp[i/2],i to even dp[i]=dp[i−1],i is odd dp[i] = Dp[i-2] + DP[I/2], I is even \ \ dp[i] = Dp[i-1], I is odd

The complexity of this is O (n) o (n)

However, the complexity of 51nod 1383 is sufficient, but for 1048 of N (1030) n (10^{30}), it is completely inadequate,

What if n is very big.
That would have to dig up the nature of the decomposition scheme.

1. For a number n, it is assumed that the binary has M 1, respectively, the a1,a2...am a_1,a_2...a_m bit. For any one of n decomposition schemes, the power of all 2 is sorted in ascending order, then can be divided into M-segment, where the sum of paragraph I is 2ai 2^{a_i}
2. For a number 2i 2^i of a partition scheme, if it is not only a number of itself, it is possible to put these 2 powers in ascending order, and then divided into two segments, each paragraph and is 2i−1 2^{i−1}

Proving not difficult. With these properties, you can set a new state.

First set g[i][j] means that the first segment is done (i.e. the first I 1 bits under n binary), the largest number is 2j 2^j, and the number of programs.
Set up an auxiliary array f[i][j], representing