MAGICHEF - Editorial

basic-programming
conditionals
easy
editorial
looping
sept18

#1

PROBLEM LINK:

Div1
Div2
Practice

Setter- Shivam Gupta
Tester- Teja Vardhan Reddy
Editorialist- Abhishek Pandey

DIFFICULTY:

CAKEWALK

PRE-REQUISITES:

Looping, Conditionals Basic Coding .

PROBLEM:

There is a gold coin under box at index X. S swaps are performed where two boxes are exchanged. We need to tell that finally, i.e. after all swaps are done, which indexed box has gold coin under it.

QUICK-EXPLANATION:

Key to AC- Having coded 5-6 problems from cakewalk question is enough to solve this problem. Looping, Conditionals and swapping of variables is needed.

We initially know that gold coin is under box X. For each swap, we know the indices (i,j) of the boxes which are being swapped. We form basic conditions of form-

  • Let Curr hold the index of box under which coin is initially. By the question, Curr=X initially. Now, for every swap, the following holds-
  • If i eq Curr and j eq Curr- Do nothing. We dont care about such swaps.
  • If i==Curr, then it means that box i and j are swapped, and coin was initially in box i. Now, after swapping, it will be under box at index j. Hence, we assign Curr=j.
  • If j==Curr then by same reasoning we assign Curr=i.
  • i==j==Curr , we need to do nothing. (I dont think this case exists in the question XD).

The final ans is in Curr. Please note I followed standard programming convention of = representing assignment operator and == representing “Check for Equality” operator.

EXPLANATION:

This is one of the editorial where Quick Explanation more than sums up everything about the logic. Instead of making editorial unnecessarily lengthy by formulating the logic again, lets focus on implementation.

In case there is any doubt of logic under Quick Explanation, do ask. I kept it short as it was simply common sense. The gist of the logic is, if the coin is currently under any one of the bowls, i or j, then it will be in the other bowl with which we swapped after the operation. Keep a track of which bowl the coin is CURRENTLY in.

Taking the input is trivial, but with that being said, let me introduce you to Fast Input-Output methods. The rule of thumb says, if number of input is >10^5 or output >10^5, use fast input-output to save precious time. Else, many times correct solutions get TLE due to huge time wasted in IO operations.

Maintain a temporary variable to store answer. I used currAns. After this, implement the conditionals mentioned above. Remember that, for cases where (i eq currAns \&\& j eq currAns) and (i==currAns \&\& j==currAns) we dont need to do nothing, hence we ignore them. We only handle the remaining 2 cases. Implementation in tab below-

Click to view
   int currAns=x;
	    int a,b;
	    while(s--)
	    {
	        cin>>a>>b;//a corresponds to i, b corresponds to j. This line takes input
	        if(a==currAns)
	            currAns=b;//if (i==currAns) currAns=j; 
	        else if(b==currAns)
	            currAns=a;//Same as above.
	    }
	    cout<<currAns<<endl;//Print

SOLUTION

The code is pasted in tabs below for you guys to refer in case solution links dont work.

Setter

Tester

Click to view
//teja349
#include <bits/stdc++.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <utility>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip> 
//setbase - cout << setbase (16); cout << 100 << endl; Prints 64
//setfill -   cout << setfill ('x') << setw (5); cout << 77 << endl; prints xxx77
//setprecision - cout << setprecision (14) << f << endl; Prints x.xxxx
//cout.precision(x)  cout<<fixed<<val;  // prints x digits after decimal in val
 
using namespace std;
#define f(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) f(i,0,n)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define vi vector< int >
#define vl vector< ll >
#define ss second
#define ff first
#define ll long long
#define pii pair< int,int >
#define pll pair< ll,ll >
#define sz(a) a.size()
#define inf (1000*1000*1000+5)
#define all(a) a.begin(),a.end()
#define tri pair<int,pii>
#define vii vector<pii>
#define vll vector<pll>
#define viii vector<tri>
#define mod (1000*1000*1000+7)
#define pqueue priority_queue< int >
#define pdqueue priority_queue< int,vi ,greater< int > >
 
// used for storing contents of boxes.   
int arr[123456];
int main(){
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
    	int n,x,s;
    	cin>>n>>x>>s;
    	int i,a,b;
        // initialising correct box
    	arr[x]=1;
        // performing swaps
    	rep(i,s){
    		cin>>a>>b;
    		swap(arr[a],arr**);
    	}
 
        // checking for answer box!!
    	rep(i,123456){
    		if(arr*){
    			arr*=0;
    			cout<<i<<endl;
    			break;
    		}
    	}
 
    }
    return 0;     
}

Editorialist

Click to view
/*
 *
 ********************************************************************************************
 * AUTHOR : Vijju123                                                                        *
 * Language: C++14                                                                          *
 * Purpose: -                                                                               *
 * IDE used: Codechef IDE.                                                                  *
 ********************************************************************************************
 *
 Comments will be included in practice problems if it helps ^^
 */
 
 
 
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
 
int main() {
	// your code goes here
	#ifdef JUDGE
    freopen("input.txt", "rt", stdin);
    freopen("output.txt", "wt", stdout);
    #endif
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	srand(time(NULL));
	int t;
	cin>>t;
	while(t--)
	{
	    int n,x,s;
	    cin>>n>>x>>s;
	    int currAns=x;
	    int a,b;
	    while(s--)
	    {
	        cin>>a>>b;
	        if(a==currAns)
	            currAns=b;
	        else if(b==currAns)
	            currAns=a;
	    }
	    cout<<currAns<<endl;
	}
	return 0;
}

Time Complexity=O(S) per test case
Space Complexity=O(1)

CHEF VIJJU’S CORNER :smiley:

1. Swapping variables here and there kind of reminds me of the std::swap function of C++ STL. Its a nice function which comes in handy for beginners and professionals alike. It can swap variables, arrays, vectors, queues…you pretty much get it XD.

2. Why did Chef bother accepting Magicians challenge if he needs us to help him? -_-

3. Related Problems: Frankly, the entire beginner section is filled to brim with such level problems. Solve any random problem from there, EXCEPT

  • Rupsa and Game
  • Chef and Weird Game.
  • One more Weird Game.

List item


#2

https://www.codechef.com/viewsolution/20123005

Please check why is it getting WA?


#3

https://www.codechef.com/viewsolution/20047736

Can someone explain why is this solution giving WA?


#4

https://www.codechef.com/viewsolution/20085700

Can someone explain why is this solution giving WA?


#5

https://www.codechef.com/viewsolution/20243312

Why does this give WA, please help, someone?


#6

https://www.codechef.com/viewsolution/21350841

can anyone tell why this is giving wrong answer


#7

Just add this after line 6:

while (t--) {

and a closing brace in the end.


#8

You have a more-than-usually magic coin. In your code, swapping two boxes means the coin always ends up under one of them, even if it was not under either to start with.


#9

For the following case, your code prints 1 instead of 3:

3 1 2 2 3 1 3


#10

You probably want to check the last element of answer instead of ans[i-1].


#11

std::cout<<x;

Ans of each test case must be printed in a new line. Make it std::cout<<x<<endl;


#12

You never produce any output, because you never enter the main loop. You missed:

cin>>n;

to read the number of test cases.