HDU 2709 Sumsets

Sumsets
Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3226    Accepted Submission(s): 1288


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

7

 

Sample Output

6

 

Source
USACO 2005 January Silver
 

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题意：问一个数划分成2的幂的和有几种分法》》》,由于方法数很大，对10^9取余；

思路：看了看别人的博客，递推得到：如果这个数是偶数，那么他等于他前面的那个数的方法数，否则等于他前面的那个数加上他的一半的那个数的方法数》》》

公式为：s[i]=s[i-1]; (偶数）

               s[i]=(s[i-1]+s[i/2])%mod;

之后打个表就可以了，复杂度O(1);

下面附上我的代码：

#include<cstdio>const int N=1e6+10;const int mod=1e9;int s[N];void solve(){s[1]=1;for(int i=2;i<=N;i++){if(i%2) s[i]=s[i-1];elses[i]=(s[i-1]+s[i/2])%mod;}}int main(){int n;solve();while(~scanf("%d",&n))printf("%d\n",s[n]);return 0;}