POJ2229Sumsets题解动态规划DP

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 7686		Accepted: 3059

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

参考NumberPyramids一题
不过此题有更好的递推做法。

状态：

d[i][j]表示前i个二的幂数凑成数j的方法数

空间可以降维到d[j]

状态转移方程：

d[j]=d[j]+d[j-c[i]]

c[i]=2^i

边界：

d[0]=1

代码：

#include<cstdio>

int d[1000005],c[25],n,i,j;
int main()
{
	scanf("%d",&n);
	c[0]=d[0]=1;
	for(i=1;i<=20;i++)
		c[i]=c[i-1]<<1;
	for(i=0;i<=20&&c[i]<=n;i++)
		for(j=c[i];j<=n;j++)
			d[j]=(d[j]+d[j-c[i]])%1000000000;
	printf("%d/n",d[n]);		
}