2015 Huawei Machine Test--integer segmentation

Title Description:

An integer can be split into the sum of 2 powers, for example:
7 = 1+ 2 + 4
7 = 1 + 2 + 2 + 2
7 = 1 + 1 + 1 + 4
7 = 1 + 1 + 1 + 2 + 2
7 = 1 + 1 + 1 + 1 + 1 + 2
7 = 1 + 1 + 1 + 1 + 1 + 1 + 1
There are six different ways to split
Another example: 4 can be split into: 4 = 4, 4 = 1+1+1+1, 4 =, 4 = 1+1+2.
use f (n) to denote the number of different splits of N, such as f (7) = 6.
required to read n (not more than 1000000), output f (n)% 1000000000.

Problem Solving Ideas:

For odd n = 2k+1, the first item of its split must be 1, consider removing this 1, in fact one by one corresponds to the split of 2k, so f (2k+1) = f (2k).
For even n = 2k: Consider a split of 1 and No 1. There is a 1 split, corresponding to the (2k-1) split one by one, the same as the above odd case: There is no 1 split, dividing each item by 2, exactly one by one corresponds to all splits of K, so f (2k) = f (2k-1) + f (k).
The end result is only required to divide the remainder by 1 billion, in the representation range of int, and therefore does not require a large number operation. Note the nature of the remainder: (a+b)%m = (a%m+b%m)%m, so as long as the remainder is taken for each intermediate result, there is no overflow problem and the final output is not changed.

The code is as follows:

public static void Main (string[] args) {Scanner sc=new Scanner (system.in), while (Sc.hasnext ()) {int input=sc.nextint (); Fenge (input);} Sc.close ();} public static void Fenge (int n) {list<integer> list=new arraylist<integer> (); list.add (0); if ((N>=1) && (n<=1000000)) {for (int i = 1; I <= n; i++) {if (i%2==1) {if (i==1) {list.add (1);} Else{list.add (List.get (i-1));}} if (i%2==0) {List.add ((List.get (i-1) +list.get (I/2))%100000000);}} System.out.println (List.get (n));} Else{system.out.println (-1);}}

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2015 Huawei Machine Test--integer segmentation