PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4
Setter: Utkarsh Gupta
Tester: Abhinav Sharma, Lavish Gupta
Editorialist: Devendra Singh
DIFFICULTY:
626
PREREQUISITES:
None
PROBLEM:
There is a contest containing 2 problems A and B.
2 strong participants P and Q participated in the contest and solved both the problems.
P made AC submissions on problems A and B at time instants P_A and P_B respectively while Q made AC submissions on problems A and B at time instants Q_A and Q_B.
It is given that the time penalty is the minimum time instant at which a participant has solved both the problems. Also the participant with the lower time penalty will have a better rank.
Determine which participant got the better rank or if there is a TIE.
EXPLANATION:
Let X represent the time when P has solved both the problems and Y represent the time when Q has solved both the problems.
X=max(P_A,P_B)
Y=max(Q_A,Q_B)
Since the time penalty is the minimum time instant at which a participant has solved both the problems and the participant with the lower time penalty will have a better rank. Hence if X<Y the participant with a better rank is P otherwise if Y<X the participant with a better rank is Q else it is a TIE.
TIME COMPLEXITY:
O(1) for each test case.
SOLUTION:
Setter's Solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
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];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
void solve()
{
int PA,PB,QA,QB;
PA=readInt(1,100,' ');
PB=readInt(1,100,' ');
QA=readInt(1,100,' ');
QB=readInt(1,100,'\n');
if(max(PA,PB)>max(QA,QB))
cout<<"Q\n";
else if(max(PA,PB)<max(QA,QB))
cout<<"P\n";
else
cout<<"TIE\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),cout.tie(NULL);
int T=readInt(1,1000,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
int pa, pb, qa, qb;
cin >> pa >> pb >> qa >> qb;
int x = max(pa, pb), y = max(qa, qb);
if (x < y)
cout << 'P' << '\n';
else if (y < x)
cout << 'Q' << '\n';
else
cout << "TIE\n";
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}