poj 2229 【完全背包dp】【递推dp】

poj 2229

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 21281		Accepted: 8281

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

题意：用2的幂次（1，2，4，8...)表示N，有多少种表示方法。

题解：

【完全背包】：

　　定义：dp[i][j]表示前i个2的幂表示j有多少种方法，则dp[i][j]=Σ{dp[i-1][j-k*c[i]]|0<k*c[i]<=N}

　　

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 const int maxn=1e6;
 const int mod=1e9;

 int f[maxn];

 int main()
 {
     int N,M;

     while(scanf("%d",&N)==)
     {
         memset(f,,sizeof(f)),f[]=;
         for(int i=;; i++)
         {
             int t=(<<(i-));
             if(t>N) break;
             for(int j=t; j<=N; j++)
             {
                 f[j]+=f[j-t];
                 while(f[j]>mod) f[j]-=mod;
             }
         }
         printf("%d\n",f[N]);
     }
     return ;
 }

【递推dp】：

　　定义：dp[N]表示答案。当N为奇数时，dp[N]=dp[N-1];当N为偶数时，若组成N的数中没有1，则把这些数除以2，就是dp[N/2]；若有1（显然有1就至少有两个1)，则去掉两个1，相当于是dp[N-2].所以dp[N]=dp[N/2]+dp[N-2]

　　

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int maxn = 1e6;
 const int mod = 1e9;

 int dp[maxn];

 int main()
 {
     int n;
     cin >> n;
     memset(dp, , sizeof(dp));
     dp[] = , dp[] = , dp[] = , dp[] = ;
     for (int i = ; i <= n; i++) {
         if (i %  == ) dp[i] = dp[i - ]%mod;
         else dp[i] = (dp[i - ] + dp[i / ]) % mod;
     }
     cout << dp[n] << endl;
     return ;
 }

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e6;
const int mod=1e9;

int dp[maxn+];

int dfs(int n)
{
    if(dp[n]>) return dp[n];
    if(n%==)
        return dp[n]=dfs(n-)%mod;
    else
        return dp[n]=(dfs(n-)+dfs(n/))%mod;
}

int main()
{
    memset(dp,,sizeof(dp));
    int n;
    dp[]=,dp[]=,dp[]=,dp[]=;
    while(scanf("%d",&n)==)
    {
        printf("%d\n",dfs(n));
    }
    return ;
}