POJ 2229 递推

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

当 i 为奇数时可以将 i-1 的展开项加 1 得到 i 的展开项
当 i 为偶数是单纯的将 i-1 的展开项加 1 无法得到所有的 i 的展开项，因为 i 是偶数，所以 i 的展开项中有全为偶数的情况
将全为偶数的展开像除以 2 得到的是 i/2 的展开项（可以倒着想，将 i/2 的展开项乘 2 得到 i 的全偶数展开项） 

转移式为 dp[i] = (i&1)?(dp[i-1]):(dp[i-1]+dp[i/2])

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;

 const int maxn = 1e6+;
 int dp[maxn];

 int main(){
     int n;
     scanf("%d",&n);
     memset(dp,,sizeof(dp));
     dp[] = ;

     for(int i=;i<=n;++i){
         if(i&)
             dp[i] = dp[i-];
         else{
             dp[i] = dp[i-] + dp[i>>];
             if(dp[i]>)
                 dp[i] -= ;
         }
     }

     printf("%d\n",dp[n]);
     return ;
 }