POJ2229-Sumsets-完全背包

最新推荐文章于 2018-07-30 08:45:32 发布

原创最新推荐文章于 2018-07-30 08:45:32 发布 · 335 阅读

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本文探讨了求和集问题，即使用2的整数次幂的数求和表示一个特定整数的所有可能方式，并提供了两种不同的算法实现方案：一种利用动态规划的方法，另一种则为完全背包问题的解决思路。

原题链接
Sumsets
Time Limit: 2000MS Memory Limit: 200000K
Total Submissions: 17986 Accepted: 7052
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+1+1+1+1+2
1+1+1+2+2
1+1+1+4
1+2+2+2
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

#include <cstdio>
#include <iostream>
using namespace std;
const int maxn = 1000000 + 10;
long long dp[maxn];
int main(){
	int n;
	cin >> n;
	dp[1]=1;
	for(int i=2;i<=n;i++){
		if(i&1) dp[i]=dp[i-1];
		else dp[i] = (dp[i/2] + dp[i-1])%1000000000;
		cout << i << ":" << dp[i] << endl;
	}
	cout << dp[n] << endl;
	return 0;
}
/*
下面是完全背包的做法
#include <cstdio>
#include <iostream>
using namespace std;
const int maxn = 1000000 + 10;
const int mod=1e9;
int dp[maxn];
int main(){
    int n;
    cin >> n;
    dp[0]=1;
    for(int i=1;i<21;i++){
        int v=(1<<(i-1));
        for(int j=v;j<=n;j++){
            dp[j] = dp[j] + dp[j-v];
            while(dp[j]>mod) dp[j]-=mod;
        }
    }
    cout << dp[n] << endl;
}
// i    0  1  2  3  4  5
// v[i] 1  2  4  8  16 32

// i\j 0  1  2  3  4  5  6  7
// 0   0  0  0  0  0  0  0  0       i:1~ceiling     j:v[i-1]~n
// 1   1  1  1  1  1  1  1  1       dp[j] = dp[j] + dp[j-v[i-1]]; v[i-1]<= j <=n;
// 2   1  1  2  2  3  3  4  4          
// 3   1  1  2  2  4  4  6  6
// 4   1  1  2  2  4  4  6  6
*/