POJ 2229 Sumsets

Sumsets

Time Limit: 2000ms

Memory Limit: 200000KB

This problem will be judged on PKU. Original ID: 2229
64-bit integer IO format: %lld      Java class name: Main

 

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

Source

USACO 2005 January Silver

 

解题：背包。。。类比下就看出来了。。。

/*
@author: Lev
@date:
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <climits>
#include <deque>
#include <sstream>
#include <fstream>
#include <bitset>
#include <iomanip>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;
const int maxn = 1000010;
int dp[maxn],c[25] = {1};
int main(){
    int x;
    for(int i = 1; i < 25; ++i)
        c[i] = c[i-1]<<1;
    while(~scanf("%d",&x)){
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        for(int i = 0;c[i] <= x; ++i){
            for(int j = c[i]; j <= x; ++j)
            dp[j] = (dp[j] + dp[j-c[i]])%1000000000;
        }
        printf("%d
",dp[x]);
    }
    return 0;
}