POJ 2229 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

Thinking 今晚POJ好像宕机了，一liu都是waiting，说实话 这还是看了别人的题解写的。虽然我看完题就知道这是完全背包，但是。。。我自己理解的太浅，完全背包写的WA了，所以还是递推吧。 太慢了，OJ还没有出结果，我要拥抱电路和离散了。结果应该是对的。PS : 22号看了一下，AC了，是正确的，嗯，加油。

#include <cstdio>#include<iostream>using namespace std;const int maxn=1000005;const long long MOD=1000000000;int dp[maxn];int main(){    int N;    while ((scanf("%d",&N))!=EOF) {        dp[1]=1;        dp[2]=2;        for (int i=3; i<=N; i++) {            if ((i%2)==1) {                dp[i]=dp[i-1];            }            else dp[i]=(dp[i-1]+dp[i/2])%MOD;        }        printf("%d\n",dp[N]);    }    return 0;}