HDU 2709 Sumsets 经典简单线性dp

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3968    Accepted Submission(s): 1578

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

 

 

Recommend

teddy

 

分析：

划分规则：
1或者2倍数

dp[i]:i的方案数

i为奇数：肯定有一个落单的1，所以就是i-1的方法数
i为偶数：两种情况
两个1：所以就是i-2的方法数
没有1：i/2的所有方案*2就是i的方案数

 

code：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

//sf???
#define max_v 1000005
LL dp[max_v];
int main()
{
    dp[1]=1;
    dp[2]=2;
    for(int i=3;i<=1000000;i++)
    {
        if(i%2==1)
            dp[i]=dp[i-1]%1000000000;
        else
            dp[i]=(dp[i-2]+dp[i/2])%1000000000;
    }
    int x;
    while(~scanf("%d",&x))
    {
        printf("%lld\n",dp[x]);
    }
    return 0;
}
/*
划分规则：
1或者2倍数

dp[i]:i的方案数

i为奇数：肯定有一个落单的1，所以就是i-1的方法数
i为偶数：两种情况
两个1：所以就是i-2的方法数
没有1：i/2的所有方案*2就是i的方案数
*/