POJ_2229_Sumsets_(动态规划)

描述

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

将一个数n分解为2的幂之和共有几种分法?

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 16207		Accepted: 6405

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

分析

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对i讨论:

1.i是奇数:

　　分成的序列中必有1,所以可将i分为1+(i-1),所以f[i]=f[i-1];

2.i是偶数:
　　(1).分成的序列中有1:

　　　　同奇数,f[i]=f[i-1];

　　(2).分成的序列中没有1:

　　　　序列中的所有数都是2的倍数,那么任一种序列中的各个数/2,就得到了i/2的序列,那这种情况下,i的序列数就和i/2的序列数相同即f[i]=f[i/2];

　　综上:f[i]=f[i-1]+f[i/2];

 #include<cstdio>

 const int maxn=,mod=1e9;
 int n,f[maxn];

 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("sum.in","r",stdin);
     freopen("sum.out","w",stdout);
 #endif
     scanf("%d",&n);
     f[]=;
     for(int i=;i<=n;i++)
     {
         if(i&) f[i]=f[i-];
         else f[i]=(f[i-]+f[i/])%mod;
     }
     printf("%d\n",f[n]);
 #ifndef ONLINE_JUDGE
     fclose(stdin);
     fclose(stdout);
 #endif
     return ;
 }