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+5;
const int Mod=1e9;
#define db double
int n;
int  dp[N];
void solve(){
    memset(dp,0,sizeof(dp));
    dp[0]=dp[1]=1;
    for(int i=0;i<22;i++){
        int v=1<<i;
        for(int j = 2;j<=N;j++){
            if(j>=v) dp[j]+=dp[j-v];
            while (dp[j]>=1000000000) dp[j]-=1000000000;
        }
    }
}
int main(){
    solve();
    while(scanf("%d",&n)==1){
        printf("%d
",dp[n]);
    }
    return 0;
}