2019牛客暑期多校训练营（第九场）D.Knapsack Cryptosystem

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

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.

题意：

n个数，一个K，问你这n个数中哪几个数的加和是n，并用1标注出来？

思路：

折半搜索

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n;
LL k,ans;
LL a[100];
int vis[100];
map<LL,LL>mp;
void dfs1(int pos,int en,LL sta,LL sum)
{

    if(pos==en+1)
    {

        if(sum==k)
        {
            ans=sta;
            return;
        }
        if(mp[sum]==0)
        {
            mp[sum]=sta;
           // cout<<sum<<" "<<sta<<endl;
        }
        return;
    }

    if(sum==k)
    {
        ans=sta;
        return;
    }
    if(mp[sum]==0)
    {
        mp[sum]=sta;
       // cout<<sum<<" "<<sta<<endl;
    }
    //cout<<sum<<" "<<sta<<endl;

    dfs1(pos+1,en,sta,sum);
    dfs1(pos+1,en,sta|(1LL<<pos),sum+a[pos]);

}
void dfs2(int pos,int en,LL sta,LL sum)
{
    
    if(pos==en+1)
    {
        if(sum==k)
        {
        
            ans=sta;
            return ;
        }
        if(mp[k-sum]!=0)
        {
            ans=mp[k-sum]+sta;
        }
        return ;
    }

    if(sum==k)
    {
        ans=sta;
        return ;
    }

    if(mp[k-sum]!=0)
    {
        ans=mp[k-sum]+sta;
    }


    dfs2(pos+1,en,sta,sum);
    dfs2(pos+1,en,sta|(1LL<<pos),sum+a[pos]);

}

int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&a[i]);
    }
    ans=0;
    int mid=(n+1)/2;

    if(ans==0)
        dfs1(1,mid,0,0);

    if(ans==0)
        dfs2(mid+1,n,0,0);


    for(int i=1; i<=n; i++)
    {
        if((ans>>i)&1)
        {
            printf("1");
        }
        else
            printf("0");
    }




    return 0;
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using namespace std;
const int N=40;
map<ll ,ll >mp;
ll index[N],a[N],target;
int n,mid;
ll ans;
void init()
{
    index[1]=1;
    for(int i=2;i<=N;i++)
        index[i]=index[i-1]*2;
}
void DFS(int pos,ll sta,ll sum,int en)
{
    if(pos==en+1)
    {
        if(pos==mid)
        {
            if(sum==target)
                ans=sta;
            mp[sum]=sta;;
        }
        else
        {
            if(mp[target-sum]!=0)
                ans=sta+mp[target-sum];
        }
        return ;
    }
    DFS(pos+1,sta,sum,en);
    DFS(pos+1,sta+index[pos],sum+a[pos],en);
}
int main()
{
    init();
    scanf("%d%lld",&n,&target);
    for(int i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    mid=n/2+1;
    ans=0;
    
    if(target) DFS(1,0,0,n/2);
    if(!ans) DFS(n/2+1,0,0,n);
    
    for(int i=1;i<=n;i++)
    {
        if(ans&1)
            cout<<"1";
        else
            cout<<"0";
        ans>>=1;
    }
    cout<<endl;
    return 0;
}