POJ 2229 Sumsets(递推)

[align=center]Sumsets[/align]

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 14964		Accepted: 5978

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：给出一个数字n，问这个n有多少种由2的次幂之和组成的形式。

题解：可以直接列举递推:

n:                                                     ans:

    1                                                          1

    2                                                          2

    3                                                          2

    4                                                          4

    5                                                          4

    6                                                          6

    7                                                          6

    8                                                          10

    9                                                          10

    10                                                        14

    11                                                        14

    12                                                         20

     ......                                                        ......

if(n&1)                                                       a
=a[n-1]

if(n%2==0)                                                a
=(a[n-2]+a[n>>1])%1000000000

由上述递推过程很容易发现递推结果，不过可能会由于列举的数字不多，只列举了前7个，误认为递推关系是               if(n&1)  a
=a[n-1];    if(n%2==0)  a
=n    弱鸡的我就是这么想的(；′⌒`)。

还有一种利用二进制递推的方式，讨论区看来的啦：

可以将n用二进制表示.

n=1,只有1种表示方法。

n=2，10(2)，二进制表示下，可以分拆成{1,1}，{10}有两种表示方法

n=3, 11(2),可以分拆成{1,1,1},{10,1}.

n=4, 100(2),{1,1,1,1},{10,1,1},{10,10},{100}.

.........

总结：如果所求的n为奇数，那么所求的分解结果中必含有1，因此，直接将n-1的分拆结果中添加一个1即可 为s[n-1]

如果所求的n为偶数，那么n的分解结果分两种情况

1.含有1 这种情况可以直接在n-2的分解结果中添加两个1即可，这种情况有 s[n-1]

2.不含有1 那么，分解因子的都是偶数，将每个分解的因子都除以2，刚好是n/2的分解结果，并且可以与之一一对应，这种情况有 s[n/2]

Attention:just one test case，代码如下：

#include<cstdio>
#include<cstring>
int a[1000010];
int main()
{
int n,i;
a[1]=1; a[2]=2;
for(i=3;i<1000001;++i)
{
if(i&1)
a[i]=a[i-1];
else
a[i]=(a[i-2]+a[i>>1])%1000000000;
}
scanf("%d",&n);
printf("%d\n",a
);
return 0;
}