【2019牛客暑期多校訓練營（第九場）】D.Knapsack Cryptosystem

題目描述

Amy asks Mr. B  problem D. Please help Mr. B to solve the following problem.

Amy wants to crack Merkle–Hellman knapsack cryptosystem. Please help it.

Given an array {ai} with length n, and the sum s.

Please find a subset of {ai}, such that the sum of the subset is s.

For more details about Merkle–Hellman knapsack cryptosystem Please read

https://en.wikipedia.org/wiki/Merkle%E2%80%93Hellman_knapsack_cryptosystem

https://blog.nowcoder.net/n/66ec16042de7421ea87619a72683f807

Because of some reason, you might not be able to open Wikipedia.

Whether you read it or not, this problem is solvable.

輸入描述:

The first line contains two integers, which are n(1 <= n <= 36) and s(0 <= s < 9 * 1018)

The second line contains n integers, which are {ai}(0 < ai < 2 * 1017).

 

{ai} is generated like in the Merkle–Hellman knapsack cryptosystem, so there exists a solution and the solution is unique.

Also, according to the algorithm,  for any subset sum s, if there exists a solution, then the solution is unique.

輸出描述:

Output a 01 sequence.

If the i-th digit is 1, then ai is in the subset.
If the i-th digit is 0, then ai is not in the subset.

示例1

輸入

8 1129
295 592 301 14 28 353 120 236

輸出

01100001

說明

This is the example in Wikipedia.

示例2

輸入

36 68719476735
1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368

輸出

111111111111111111111111111111111111

直接爆搜複雜度是O(2^n),最大爲2的36次方，不可取

採用折半搜索的方法，先搜索前半段，找出所有組合的結果，然後搜索後半段，判斷搜索的結果能否組成正確的和

代碼：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))

ll n,s,a[100005];
string now;
map<ll,string>mp;
map<ll,int>f;

void dfs(int step,string k,ll sum)
{
	if(step==n/2)
	{
		f[sum]=1;
		mp[sum]=k;
		return ;
	}
	dfs(step+1,k+'0',sum);
	dfs(step+1,k+'1',sum+a[step]);
}

void dfs1(int step,string k,ll sum)
{
	if(step==n)
	{
		if(f[s-sum])
		{
			cout<<mp[s-sum]<<k<<endl;
			exit(0);
		}
		return ;
	}
	dfs1(step+1,k+'0',sum);
	dfs1(step+1,k+'1',sum+a[step]);
}

int main()
{
	int i;
	cin>>n>>s;
	for(i=0;i<n;i++)
		cin>>a[i];
	dfs(0,now,0);
	dfs1(n/2,now,0);
	return 0;
}