hdoj--2709--Sumsets(数位dp)

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Sumsets

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1996 Accepted Submission(s): 786

Problem 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

Sample Output

Source

USACO 2005 January Silver

/*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.解题思路：1.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-1的分解结果中添加一个1即可 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]*//*#include<stdio.h>#include<string.h>int a[1000010];int main(){    a[1]=1;    a[2]=2;    int n,i=3;    while(i<=1000000){            a[i++]=a[i-1];            a[i++]=(a[i-1]+a[i>>1])%1000000000;    }    while(scanf("%d",&n)!=EOF){        printf("%d\n",a[n]);    }    return 0;}*/#include<stdio.h>#include<string.h>int f[1000010];int main(){int n;while(scanf("%d",&n)!=EOF){memset(f,0,sizeof(f));f[1]=1;for(int i=2;i<=n;i++){if(i&1)f[i]=f[i-1];elsef[i]=(f[i-1]+f[i>>1])%1000000000;}printf("%d\n",f[n]);}return 0;}

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