2019牛客暑期多校训练营（第九场）D:Knapsack Cryptosystem（折半搜索）

本文深入探讨了Merkle-Hellman背包密码系统的破解方法，通过详细的算法描述和代码实现，展示了如何找到特定子集使得其元素之和等于给定的目标值。文章提供了一个具体的示例，帮助读者理解背包密码系统的工作原理。

链接：https://ac.nowcoder.com/acm/contest/889/D

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

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.

输入描述:

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.

输出描述:

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

输入

复制

8 1129
295 592 301 14 28 353 120 236

输出

复制

01100001

说明

This is the example in Wikipedia.

示例2

输入

复制

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

输出

复制

111111111111111111111111111111111111

n最多为36，直接dfs，需要 $2^{36}$ ，显然不行

可以先搜索前一半，并记录每种状态对应的结果

在搜索后一半时，若存在和相加等于s，则输出

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;

#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define continue(x) { x; continue; }
#define break(x) { x; break; }
const int N = 3e5 + 10;
int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
int a[50];
ll b[50];
int record[N][20];
ll s;
int n;
int flag;
unordered_map<ll, vector<int>>mp;
void dfs1(int d, ll sum)
{
	//if (sum == 262143)
	//	cout << "Y" << endl;
	if (d == n / 2 + 1)
	{
		if (mp.find(sum) != mp.end())
			return;
		for (int i = 1; i <= n / 2; i++)
			mp[sum].push_back(a[i]);
		return;
	}
	else
		for (int i = 0; i < 2; i++)
		{
			a[d] = i;
			if (i == 1)
				dfs1(d + 1, sum + b[d]);
			else
				dfs1(d + 1, sum);
			a[d] = 0;
		}
}
void dfs2(int d, ll sum)
{
	if (flag || d > n + 1)
		return;
	if (mp.find(s - sum) != mp.end()) // 如果找到了
	{
		ll x = s - sum;
		int len = mp[x].size();
		for (auto &i : mp[x])
			cout << i;
		for (int i = n / 2 + 1; i <= n; i++)
			cout << a[i];
		flag = 1;
		return;
	}
	else
	{
		for (int i = 0; i < 2; i++)
		{
			a[d] = i;
			if (i == 1)
				dfs2(d + 1, sum + b[d]);
			else
				dfs2(d + 1, sum);
			a[d] = 0;
		}
	}
}
int main(){
#ifdef LOCAL
	freopen("D:/input.txt", "r", stdin);
#endif
	ll sum = 0;
	cin >> n >> s;
	for (int i = 1; i <= n; i++)
		scanf("%lld", &b[i]), sum += b[i];
	dfs1(1, 0);
	dfs2(n / 2 + 1, 0);
	return TIME;
}