POJ 2229 动态规划

POJ 2229 动态规划问题

题目的链接如下：

http://poj.org/problem?id=2229

Sumsets

Time Limit: 2000MS Memory Limit: 200000K

Total Submissions: 16603 Accepted: 6574

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

//贼简单，不加注释了，完全背包再加；
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
const int MAX = 1e9;
/*int V;
int f[10005]; // 当前的状态，就是最小值
int w[10005]; // 物品的价值，就是
int v[10005]; // 物品的体积，也就是金币的重量*/
int dp[1000005];
using namespace std;
int main()
{
int n, m, k; // V就是总体积，总金额
while(cin >> n){
dp[1] = 1;
for(int i = 2 ; i <= n ; i++)
if(i % 2 == 1)
dp[i] = dp[i-1] % MAX;
else
dp[i] = (dp[i-1] + dp[i/2]) % MAX ;
cout << dp
<< endl;
}
return 0;
}