poj 2229 Sumsets（dp 或 数学）

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 . Here are the possible sets of numbers that sum to : 

) ++++++
) +++++
) ++++
) +++
) +++
) ++ 

Help FJ count all possible representations for a given integer N ( <= N <= ,,). 

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  digits (in base  representation).

Sample Input

Sample Output

Source

USACO 2005 January Silver

 

如果i为奇数，肯定有一个1，把f[i-1]的每一种情况加一个1就得到fi,所以f[i]=f[i-1]

如果i为偶数，如果有1，至少有两个，则f[i-2]的每一种情况加两个1，就得到i，如果没有1，则把分解式中的每一项除2，则得到f[i/2]

所以f[i]=f[i-2]+f[i/2]

由于只要输出最后9个数位，别忘记模1000000000

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<cmath>
 using namespace std;
 #define N 1010006
 #define MOD 1000000000
 int n;
 int dp[N];
 void init(){

     dp[]=;
     dp[]=;
     for(int i=;i<N;i++){
         if(i&){
             dp[i]=dp[i-];
         }
         else{
             dp[i]=dp[i-]+dp[i/];
             dp[i]%=MOD;
         }
     }
 }
 int main()
 {
     init();
     while(scanf("%d",&n)==){

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

     }
     return ;
 }