HDU 2709 Sumsets（递推）

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Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2814    Accepted Submission(s): 1124

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

就是把一个数分成2的n次方，问一共有多少种分法。

写出前10项之后，偶然就发现了规律，如果是奇数，就等于他前面的那个数，如果是偶数，就等于a[i-1]+a[i/2]。注意要模除1000000000.

代码实现：

#include<iostream>#include<algorithm>#include<cmath>using namespace std;double a[1000005];int main(){int i,j,k,n;a[1]=1;a[2]=2;for(i=3;i<=1000000;i++){if(i%2==1)a[i]=a[i-1];elsea[i]=(int)(a[i-1]+a[i/2])%1000000000;}while(cin>>n){printf("%.0lf\n",a[n]);}return 0;}