bzoj 1677: [Usaco2005 Jan]Sumsets 求和-布布扣-bubuko.com

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).

给出一个N(1≤N≤10^6)，使用一些2的若干次幂的数相加来求之．问有多少种方法

Input

一个整数N.

Output

方法数．这个数可能很大，请输出其在十进制下的最后9位．

Sample Input

Sample Output

6

有以下六种方式
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

HINT

Source

Silver

#include<cstdio>
#define mod 1000000000
#define F(i,l,r) for(int i=l;i<=r;i++)
int dp[1000010],n;
int main()
{
	dp[1]=1;
	scanf("%d",&n);
	F(i,2,n)
	{
		if(i&1) dp[i]=dp[i-1];
		else dp[i]=(dp[i-1]+dp[i>>1])%mod;
	}
	printf("%d",dp[n]);
}

bzoj 1677: [Usaco2005 Jan]Sumsets 求和

原文：http://www.cnblogs.com/lkhll/p/5854175.html