hdu 2709（递推）

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2558    Accepted Submission(s): 1031

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

1.n为奇数，a[n]=a[n-1]

2.n为偶数：

（1）如果加数里含1，则一定至少有两个1，即对n-2的每一个加数式后面 +1+1，总类数为a[n-2]；

（2）如果加数里没有1，即对n/2的每一个加数式乘以2，总类数为a[n-2]；

所以总的种类数为：a[n]=a[n-2]+a[n/2];

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int a[1000005];int main(){a[1]=1;a[2]=2;for(int i=3;i<1000005;i++){if(i%2==0){a[i]=(a[i-2]+a[i/2])%1000000000;}elsea[i]=a[i-1];}int n;while(~scanf("%d",&n)){printf("%d\n",a[n]);}return 0;}