poj 2229 动态规划

怼烎@ 2022-08-03 14:47 75阅读 0赞

**Sumsets**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 200000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 14447</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 5763</td> 
  </tr> 
 </tbody> 
</table>

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

Sample Output

Source

[USACO 2005 January Silver][]

仔细观察数列就会发现，如果n是奇数的话，肯定至少有一个1，去掉每个数列中的一个1，这不就是前一个数的数列个数吗。

所以当n是奇数的时候，划分数是等于n的前一个的。

再来看偶数的情况，偶数有一些存在1，有一些不存在1，可以分成两部分来看啊，

存在1的那一些，不就等于上一个数（必定是奇数）的每个数列都加上个1吗。

不存在1的那一些，肯定都是2的倍数，因此先把他们都除二，这不就是n/2的划分数吗。

#include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;
        while((c=getchar())<'0' || c>'9');
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res;
    }
    int dp[1000010];
    const int mod=1000000000;
    int main()
    {
        dp[1]=1;
        for(int i=2;i<=1000000;i++)
        {
            if(i&1)dp[i]=dp[i-1];
            else dp[i]=(dp[i-1]+dp[i>>1])%mod;
        }
        int n;
        while(~scanf("%d",&n))
        {
            printf("%d\n",dp[n]);
        }
        return 0;
    }

[USACO 2005 January Silver]: http://poj.org/searchproblem?field=source&key=USACO+2005+January+Silver