POJ2229 递推

Sumsets

Time Limit: 2000MS Memory Limit: 200000K

Total Submissions: 19030 Accepted: 7436

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

解题思路

因为每个数量为2的整数幂 所以 当n=基数的时候 dp
=dp[n-1];

先写出前面的几个数字的种数

dp[1]=1;

dp[2]=2;

dp[3]=2;

dp[4]=4;

dp[5]=4;

dp[6]=6;

dp[7]=6;

不难发现规律 n为基数时 dp
=dp[n-1] n位偶数时 dp
=dp[n-2]+dp[n-2];

AC代码

import java.util.Scanner;

public class Main{

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan=new Scanner(System.in);

int dp[]=new int[1000005];
dp[1]=1;
dp[2]=2;
for(int i=3;i<=1000000;i++){
if(i%2!=0){
dp[i]=dp[i-1];
}
else{
dp[i]=dp[i-2]+dp[i/2];
}
if(dp[i]>1000000000){
dp[i]%=1000000000;
}
}
int n=scan.nextInt();
System.out.println(dp
);

}

}