HDU 2709 Sumsets —— 递推

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2709

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3094    Accepted Submission(s): 1225

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

题解：

此题的关键是以“1”作为讨论对象。

1.当n为奇数时，即比n-1多了一个1，且至少有一个1,。那么相当于在n-1的所有情况中，在加多一个1。所以：dp[n] = dp[n-1]

2.当n为偶数时：如果有1，那么至少有两个1，这就相当于在dp[n-2]的所有情况下加多两个1，所以dp[n] += dp[n-2]；如果没有1，那么每一项都是2的倍数，即dp[n/2]每种情况的每一项都乘上2，所以dp[n] += dp[n/2]。综上：dp[n] = dp[n-2] + dp[n/2]。

代码如下：

#include <bits/stdc++.h>#define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)#define ms(a,b) memset((a),(b),sizeof((a)))using namespace std;typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const double eps = 1e-6;const int mod = 1e9;const int maxn = 1e6+10;LL n,dp[maxn];int main(){    dp[0] = 1;    rep(i,1,maxn-1)    {        if(i%2)            dp[i] = dp[i-1];        else            dp[i] = (dp[i-2] + dp[i/2])%mod;    }    while(cin>>n)        cout<<dp[n]<<endl;}