PASSORFAIL - Editorial

PROBLEM LINK:

Practice
Div1
Div2
Div3

Setter: Utkasrsh Gupta
Tester: Manan Grover
Editorialist: Ajit Sharma Kasturi

DIFFICULTY:

CAKEWALK

PREREQUISITES:

None

PROBLEM:

Chef writes and exam consisting of N questions. Chef gets 3 marks for every correct answer and -1 marks for every incorrect answer.The passing marks is P for this course. Chef has answered X questions correct and the remaining questions incorrect. We need to find whether chef will pass the exam or not.

EXPLANATION:

  • Number of questions answered correctly is X. Therefore, number of questions answered incorrectly is N-X.

  • The number of marks the chef will get will then be 3\cdot X - (N-X).

  • Since the passing marks is P, chef needs atleast P marks to pass.

  • Therefore, if 3\cdot X - (N-X) \geq P we ouput PASS, else we output FAIL.

TIME COMPLEXITY:

O(1) for each testcase.

SOLUTION:

Editorialist's solution
#include <bits/stdc++.h>
using namespace std;

int main()
{
     int tests;
     cin >> tests;
     while (tests--)
     {
          int n, x, p;
          cin >> n >> x >> p;

          if (3 * x - (n - x) >= p)
               cout << "PASS" << endl;
          else
               cout << "FAIL" << endl;
     }
}


Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val)  no. of elements strictly less than val
// s.find_by_order(i)  itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
void solve()
{
    ll n,x,p;
    cin>>n>>x>>p;
    assert(n>=1 && n<=100);
    assert(x>=0 && x<=n);
    assert(p>=0 && p<=3*n);
    ll marks=3*x-(n-x);
    if(marks>=p)
        cout<<"PASS\n";
    else
        cout<<"FAIL\n";
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T=1;
    cin>>T;
    assert(T>=1 && T<=1000);
    int t=0;
    while(t++<T)
    {
        //cout<<"Case #"<<t<<":"<<' ';
        solve();
        //cout<<'\n';
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}

Tester's solution
#include <bits/stdc++.h>
using namespace std;
int main(){
  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
  int t;
  cin>>t;
  while(t--){
    int n,x,p;
    cin>>n>>x>>p;
    if(p <= 4 * x - n){
      cout<<"PASS\n";
    }else{
      cout<<"FAIL\n";
    }
  }
  return 0;
}


Please comment below if you have any questions, alternate solutions, or suggestions. :slight_smile: