Coin Ecryption SOLUTIONS BINCOIN

Coin Ecryption SOLUTIONS BINCOIN

Culinary expert and Ada have been alloted an errand of finding a mystery paired code which has been encoded in a strange way. There are N and M coins covered up in the ways followed by Chef and Ada individually. Each coin has a number composed on one side while opposite side contains a letters in order. Gourmet expert and Ada begin finding the coins from the beginning stages of their particular ways. In this cycle they probably won’t have the option to discover a few coins and push forward. 

There is an uncommon possibility that the all the coins picked and the request for picking them will be comparative for Chef and Ada regarding numbers composed on them. Let the most number of coins that these two can get and fulfill this uncommon condition be K. The mystery double code is the twofold portrayal of XOR of ASCII portrayals of the letters in order composed on coins at area K on the ways of Chef and Ada separately. 

 

Information 

The main line of the info contains a solitary number T meaning the quantity of experiments. The portrayal of T experiments follows. 

The primary line of each experiment contains two space-isolated whole numbers N and M. 

The second line of each experiment contains N space isolated numbers C1,C2,C3,… CN. 

The third line of each experiment contains a line of length N. 

The fourth line of each experiment contains M space isolated numbers A1,A2,A3,… AN. 

The fifth line of each experiment contains a line of length M. 

Note: The letters in order at position I in the string compares to the whole number at position I in the succession. 

 

Yield: 

For each expeiment print a parallel string – double portrayal of XOR of ASCII estimations of letter sets at area K. 

Note : Output of each experiment ought to be imprinted on new line. 

 

Limitations 

1≤T≤102 

1≤N≤103 

1≤M≤103 

1≤c[i],a[i]≤109 

 

Model Input 

3 4 

1 2 3 

abc 

2 3 4 5 

pqrs 

 

Model Output 

10011 

 

Clarification: 

In the best case Chef and Ada can share coins [2,3] for all intents and purpose for example 2 comparable coins subsequently the parallel code is the XOR of ASCII estimation of second component of both the strings for example ASCII estimation of b = 98, q = 113 consequently we will find our last solution by doing 98 XOR 113.

 

 

March Long Challenge 2021 Solutions

April Long Challenge 2021 Solutions

Codechef Long Challenge Solutions

February Long Challenge 2021

1. Frog Sort Solution Codechef

2. Chef and Meetings Solution Codechef

3. Maximise Function Solution Codechef

4. Highest Divisor Solution Codechef

5. Cut the Cake Challenge Solution Codechef

6. Dream and the Multiverse Solution Codechef

7. Cell Shell Solution Codechef

8. Multiple Games Solution Codechef

9. Another Tree with Number Theory Solution Codechef

10. XOR Sums Solution Codechef

11. Prime Game Solution CodeChef

12. Team Name Solution Codechef

January Long Challenge 2021

November Challenge 2020 SOLUTION CodeChef

October Lunchtime 2020 CodeChef SOLUTIONS

RELATED :

Related :

Related :

1 thought on “Coin Ecryption SOLUTIONS BINCOIN”

  1. #include
    using namespace std;
    #define int long long
    #define rapido ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    #define endl "n"

    int lcs(vector&arr1,vector&arr2,int n,int m)
    {
    int dp[n+1][m+1];
    for (int i = 0; i <= n; i++)
    {
    for (int j = 0; j <= m; j++)
    {
    if (i == 0 || j == 0)
    dp[i][j] = 0;

    else if (arr1[i – 1] == arr2[j – 1])
    dp[i][j] = dp[i – 1][j – 1] + 1;

    else
    dp[i][j] = max(dp[i – 1][j], dp[i][j – 1]);
    }
    }
    return dp[n][m];
    }

    void solve()
    {
    int n;int m;cin>>n>>m;
    mapmp1;
    mapmp2;
    vectorarr1(n);
    vectorarr2(m);
    string s1;
    string s2;

    for(int i=0;i>arr1[i];
    }
    cin>>s1;
    for(int i=0;i>arr2[i];
    }
    cin>>s2;

    int c=lcs(arr1,arr2,n,m);
    //cout<0)
    {
    int d=ans%2;
    if(d==0)
    res+="0";
    else
    res+="1";
    ans=ans/2;
    }
    reverse(res.begin(),res.end());
    cout<>t;
    while(t–)
    solve();
    return 0;
    }

    Reply

Leave a Comment