HDU 2709 Sumsets 递推

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2489    Accepted Submission(s): 1003

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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设a[n]为和为 n 的种类数；
根据题目可知，加数为2的N次方，即 n 为奇数时等于它前一个数 n-1 的种类数 a[n-1] ，若 n 为偶数时分加数中有无 1 讨论，即关键是对 n 为偶数时进行讨论：
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 <stdio.h>long long a[1000001]={0,1,2};int main(){int i,t;for(i=3;i<1000001;i++)    {        if(i%2==0)            a[i]=a[i-2]+a[i/2];        else            a[i]=a[i-1];         a[i]%=1000000000;       }while(~scanf("%I64d",&t)){printf("%I64d\n",a[t]);}return 0;}