poj 2229 Sumsets（dp 或 数学）

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

 

如果i为奇数，肯定有一个1，把f[i-1]的每一种情况加一个1就得到fi,所以f[i]=f[i-1]

如果i为偶数，如果有1，至少有两个，则f[i-2]的每一种情况加两个1，就得到i，如果没有1，则把分解式中的每一项除2，则得到f[i/2]

所以f[i]=f[i-2]+f[i/2]

由于只要输出最后9个数位，别忘记模1000000000

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
#define N 1010006
#define MOD 1000000000
int n;
int dp[N];
void init(){
    
    dp[1]=1;
    dp[2]=2;
    for(int i=3;i<N;i++){
        if(i&1){
            dp[i]=dp[i-1];
        }
        else{
            dp[i]=dp[i-1]+dp[i/2];
            dp[i]%=MOD;
        }
    }
}
int main()
{
    init();
    while(scanf("%d",&n)==1){
        
        printf("%d
",dp[n]);
    
    }
    return 0;
}