hdu 2709：Sumsets（递推）

http://acm.hdu.edu.cn/showproblem.php?pid=2709

Problem 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

题意分析：

只使用整数幂为2的数字拼凑数字，求一个有多少种可能。

解题思路：

一个奇数，可能数和比它小1的数相同，

比如9和8，把它分成1和8,8那一部分就是a[8],而这边的1不能再操作。

一个偶数8分成1和77那一部分是a[7],再加上一个1

从1 1 2 2 2 开始，变成2 2 2 2，后面的情况就像是4个1来组成4一样，有a[4]种，即a[i/2]。

所以有公式i为奇数时:  a[i]=a[i-1];

                  i为偶数时:  a[i]=a[i-1]+a[i/2];     

#include <stdio.h>
#define N 1000020
int a[N];
int main()
{
	int n;
	a[1]=1;
	for(int i=2; i<N; i++)
		if(i%2==1)
			a[i]=a[i-1];
		else
			a[i]=(a[i-1]+a[i/2])%1000000000; 
	while(scanf("%d", &n)!=EOF)
		printf("%d
", a[n]);
	return 0;
}