POJ2229 HDU2709 Sumsets【數列】

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 23695		Accepted: 9112

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

 

問題鏈接：POJ2229 HDU2709 Sumsets

問題分析：

  這個問題與OEIS中的一個數列相關聯，數列的序號是A018819。該數列的名字就是Binary partition function: number of partitions of n into powers of 2。數列問題的關鍵是該數列的通項公式，該數列的通項公式是條件遞推式。有關該數列的通項公式，參見程序。

  這個問題打表是必要的，用函數init()來實現。

程序說明：（無）

題記：（略）

參考鏈接：（無）

 

AC的C++語言程序如下：

/* POJ2229 HDU2709 Sumsets */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 1e6;
const int MOD = 1e9;
int f[N + 1];

void init()
{
    f[0] = 1;
    f[1] = 1;
    for(int i = 2; i <= N; i++)
        if(i % 2)
            f[i] = f[i - 1];
        else
            f[i] = (f[i - 1] + f[i >> 1]) % MOD;
}

int main()
{
    init();

    int n;
    while(~scanf("%d", &n))
        printf("%d\n", f[n]);

    return 0;
}