Sumsets

 

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

//递推
 题意:给出一个整数n,求解该整数n有多少种由2的幂次之和组成的方案.
    这里将n用二进制表示,当n为奇数时,n-1必定是偶数,n是在n-1的二进制的最低位加1而来,所以num[i]=num[i-1]
    而当n为偶数时,则可以划分成前面有1组成而后两位必是0的二进制表示和两部分都表示偶数的二进制,即num[i]=num[i-2]+num[i/2].
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int e[1000005];
int main(){
	int i;
	e[1]=1;
	e[2]=2;
	for(i=3;i<=1000000;i++){
		if(i%2==1)
		e[i]=e[i-1];
		else 
		e[i]=(e[i-2]+e[i/2])%1000000000;
	}
	int n;
	while(~scanf("%d",&n)){
		printf("%d\n",e[n]);
	}
	return 0;
}
//dp
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
#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]);          
}