poj 2229 DP

Sumsets

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

思路：用完全背包的方法做，状态转移方程：dp[j]+=dp[j-v](j>=v)
代码：

 #include "cstdio"
 #include "algorithm"
 #include "cstring"
 #include "queue"
 #include "cmath"
 #include "vector"
 #include "map"
 #include "stdlib.h"
 #include "set"
 typedef long long ll;
 using  namespace std;
 const int N=1e6+;
 const int Mod=1e9;
 #define db double
 int n;
 int  dp[N];
 void solve(){
     memset(dp,,sizeof(dp));
     dp[]=dp[]=;
     for(int i=;i<;i++){
         int v=<<i;
         for(int j = ;j<=N;j++){
             if(j>=v) dp[j]+=dp[j-v];
             while (dp[j]>=) dp[j]-=;
         }
     }
 }
 int main(){
     solve();
     while(scanf("%d",&n)==){
         printf("%d\n",dp[n]);
     }
     return ;
 }