POJ 2229 Sumsets（简单DP）

Posted 2020-10-06 Farewell

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
题意：
　　
　　一个数N能被拆分成多少种只由2的n次方相加的组合。
题解：
　　
　　刚开始怎么也推不出来，╮(╯▽╰)╭自己还是好菜啊。。看了别人的题解后发现其实很容易。？
　　当i为奇数时，dp[i]=dp[i-1];当i为偶数时，dp[i]=dp[i-1]+dp[i/2]。
     i为奇数，dp[i]等于前一个偶数的组合+1。偶数时，等于前一个数+1的组合数，加上dp[i/2]的组合数（可以看成提个公因子2）。

#include<iostream>
using namespace std;
const int maxn=1e6+5,mod=1e9;
int dp[maxn];
int main()
{
    int n;
    cin>>n;
    dp[1]=1,dp[2]=2;
    for(int i=3;i<=n;i++)
        dp[i]=(i&1)?dp[i-1]:(dp[i-2]+dp[i>>1])%mod;
    cout<<dp[n]<<endl;
}