Sumsets（数学）

Sumsets

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

简单的找规律的题，算不上DP

#include<cstdio>
const int maxn = 1000000 + 1;
int a[maxn];

int main()
{
	int n;	
	while(scanf("%d",&n)!=EOF)
	{
		int i,j;
		a[1] = 1;
		a[2] = 2;
		for(i=3; i<=maxn; i++)
		{
			if(i%2 == 1)
				a[i] = a[i-1];
			else
			{
				a[i] = a[i-2] + a[i/2];
				a[i] = a[i]%1000000000;	//？ 
			}
		}		
		printf("%d
",a[n]);
	}
	return 0;
} 

　　

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[n]=a[n-1]

if(n%2==0)                                                a[n]=(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[n]=a[n-1];    if(n%2==0)  a[n]=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]