POJ 2229 Sumsets (递推&整数划分变形)

Sumsets

http://poj.org/problem?id=2229

Time Limit: 2000 MS     

Memory Limit: 200000 KB

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

Source

USACO 2005 January Silver

思路：假设加数按从小到大的顺序。当n为奇数时，第一个数必须为1，此时f(n)=f(n-1)；当n为偶数时，分两种情况讨论，若第一个数为1，则f(n)=f(n-1)，若第一个数不为奇数，则所有数都不为奇数，提出一个公因子2出来，就是f(n/2)，所以，f(n)=f(n-1)+f(n/2)

完整代码：

/*63ms,4300KB*/

#include<cstdio>
const int mod = 1e9;
const int maxn = 1000001;

int f[maxn];

int main()
{
	int n, i;
	f[1] = 1, f[2] = 2;
	for (i = 3; i < maxn; ++i)
		f[i] = (i & 1 ? f[i - 1] : (f[i - 2] + f[i >> 1]) % mod) ;
	while (~scanf("%d", &n))
		printf("%d\n", f[n]);
	return 0;
}