stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <iostream>

#include<bits/stdc++.h>

using namespace std;





void countStr(char ser[], char* arr[],int n)

{

    int coun =0;



    for(int i=0 ; i<n; i++)

    {





        if(strcmp(arr[i],ser) ==0)

        {

            coun++;

        }



    }

    cout<<ser<<" "<<"x"<<" "<<coun<<endl;



}



int main()

{

    int n;

    cin>>n;

    char* arr[n];

    char x[3]="AC";





    if(n>0 && n<10^5)

    {

        cin.ignore();

        for(int i=0 ; i<n; i++)

        {

            arr[i]= new char[4];





            cin.getline(arr[i],4);



            arr[i][4]='\n';

            // cout<<arr[i]<<endl;

            if(strcmp(arr[i],x)!=0   &&  strcmp(arr[i],(char*)"WA")!=0    &&  strcmp(arr[i],(char*)"TLE")!=0   &&  strcmp(arr[i],(char*)"RE")!=0)

            {



                return 0;

            }





        }



        countStr("AC", arr,n);

        countStr("WA", arr,n);

        countStr("TLE",arr,n);

        countStr("RE", arr,n);





    }





    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.244021119348664x):
#include <stdio.h>



int main() {



	int n;

	scanf("%d", &n);

	char arr[5];

	

	int sumA = 0, sumW = 0, sumT = 0, sumR = 0;

	for(int i = 0; i<n; i++){

		scanf("%s", arr);

		if(arr[0]=='A'){

			sumA++;

		}

		else if(arr[0]=='W'){

			sumW++;

		}

		else if(arr[0]=='T'){

			sumT++;

		}

		else if(arr[0]=='R'){

			sumR++;

		}

	}



	printf("AC x %d\n", sumA);

	printf("WA x %d\n", sumW);

	printf("TLE x %d\n", sumT);

	printf("RE x %d\n", sumR);



    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <algorithm>

#include <cstring>

#include <string>

#include <cmath>

#include <set>

#include <unordered_set>

#include <vector>

#include<valarray>

#include<map>

#include <bits/stdc++.h>

#include <utility>

#include<bits/stdc++.h>

int a[4];

using namespace std;



int main()

{

int t;

cin>>t;

string s;

while (t--)

{

    cin>>s;

    if(s=="AC")

    {

        a[0]++;

    }

    if(s=="WA")

    {

        a[1]++;

    }

    if(s=="TLE")

    {

        a[2]++;

    }

    if(s=="RE")

    {

        a[3]++;

    }



}

cout<<"AC x "<<a[0]<<endl;

    cout<<"WA x "<<a[1]<<endl;

    cout<<"TLE x "<<a[2]<<endl;

    cout<<"RE x "<<a[3]<<endl;





    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.254136521891219x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main()

{

	int t, wa=0 ,tle=0, ac=0 ,re=0; cin>>t;



	for(int i=0; i< t;++i)

	{

		string s; cin>>s;

		if(s == "AC") ac++;

		else if(s == "RE") re++;

		else if(s == "TLE") tle++;

		else wa++;



	}

	cout<<"AC x "<<ac<<endl;

	cout<<"WA x "<<wa<<endl;

	cout<<"TLE x "<<tle<<endl;

	cout<<"RE x "<<re<<endl;









}

// Faster C++ solution for the same problem (measured speedup: 5.2518149675272285x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cmath>

#include <algorithm>

#include <cstdlib>

using namespace std;



int main() {

    int x, a=0,b=0,c=0,d=0;

    scanf("%d",&x);

    string s;

    for (int i = 0; i < x; i++) {

        cin>>s;

        if(s=="AC")

            a++;

        else if (s=="WA")

        b++;

        else if(s=="TLE")

            c++;

        else if(s=="RE")

            d++;

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n",a,b,c,d);



}

// Faster C++ solution for the same problem (measured speedup: 5.202911098720898x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

/*  Writen by : AS7

 *  "JUST KEEP GOING..."

 */

#define endl			'\n'

#define ll				long long

#define all(v)			((v).begin()), ((v).end())

#define allr(v)			((v).rbegin()), ((v).rend())

#define sz(v)			((int)((v).size()))

#define clr(v, d)		memset(v, d, sizeof(v))

#define rep(i, v)		for(int i=0;i<sz(v);++i)

#define For(i,s)        for(auto i:s)

#define Forr(i,s)        for(auto &i:s)

#define lp(i, n)		for(int i=0;i<(int)(n);++i)

#define lpi(i, j, n)	for(int i=(j);i<(int)(n);++i)

#define lpd(i, j, n)	for(int i=(j);i>=(int)(n);--i)

#define vi				vector <int>

#define vl				vector <ll>

#define ii				pair <int,int>

#define vii				vector <ii>

#define vll				vector <pair<ll,ll>>

#define PI				double PPPPPP = 3.14159265358979323846;

#define bit(mask,i) ((mask>>i)&1)

const int N = 2e5+5;

map<string ,int > freq;

void solve(){

   int n;

   cin>>n;

   lp(i,n){

       string s;

       cin>>s;

       freq[s]++;

   }

   cout<<"AC x "<<freq["AC"]<<"\n"<<"WA x "<<freq["WA"]<<"\n"<<"TLE x "<<freq["TLE"]<<"\n"<<"RE x "<<freq["RE"]<<endl;



}

int main() {

    int t = 1;

  //  cin>>t;

    while (t--)solve();

}

// Faster C++ solution for the same problem (measured speedup: 5.267604436125608x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
# include <iostream>

using namespace std;

//AC, WA, TLE, or RE.

int main()

{

    string a[] = {"AC", "WA", "TLE", "RE"};

    int res[] = {0,0,0,0};

    int x;

    cin >> x;

    string n;

    for(int i = 0; i < x; i++)

    {

        cin >> n;

        for (int j = 0; j < 4; j++)

        {

            if (a[j] == n)

            {

                res[j]++;

            }

        }

    }

    for (int i = 0; i < 4; i++)

    {

        cout << a[i] << " x " << res[i] << endl; 

    }

}


// Faster C++ solution for the same problem (measured speedup: 5.24419949616714x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

int main() {

    string AC , WA, TLE, RE;

    string x;

    int n;

    cin>>n;

    int cnt1=0;

    int cnt2=0;

    int cnt3=0;

    int cnt4=0;

    for (int i = 0; i <n ; ++i) {

        cin>>x;

        if(x== "AC"){++cnt1;}

        if(x== "WA"){++cnt2;}

        if(x== "TLE"){++cnt3;}

        if(x== "RE"){++cnt4;}

    }

    cout<<"AC"<<" "<<"x"<<" "<<cnt1<<endl;

    cout<<"WA"<<" "<<"x"<<" "<< cnt2<<endl;

    cout<<"TLE"<<" "<<"x"<<" "<< cnt3<<endl;

    cout<<"RE"<<" "<<"x"<<" "<< cnt4<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.257793030408738x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <iostream>

#include<bits/stdc++.h>



using namespace std;



int main()

{

    int n;

    cin>>n;

    map<string,int>tmp;





    tmp["AC"]=0;

    tmp["WA"]=0;

    tmp["TLE"]=0;

    tmp["RE"]=0;



    for(int i = 0 ; i  < n ;i++)

    {

        string s;cin>>s;



        tmp[s]++;

    }



    cout<<"AC x " << tmp["AC"]<<endl;

    cout<<"WA x " << tmp["WA"]<<endl;

    cout<<"TLE x " << tmp["TLE"]<<endl;

    cout<<"RE x " << tmp["RE"]<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.274039079417935x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



int main()

{



    int n;

    cin>>n;

    int a[4]={0};

    for(int i=0 ; i<n ; i++){

        string s;

        cin>>s;

        if(s=="AC")

            a[0]+=1;

        else if(s=="WA")

            a[1]+=1;

        else if(s=="TLE")

            a[2]+=1;

        else

            a[3]+=1;

    }





     cout<<"AC x "<<a[0]<<endl;

     cout<<"WA x "<<a[1]<<endl;

     cout<<"TLE x "<<a[2]<<endl;

     cout<<"RE x "<<a[3]<<endl;















}


// Faster C++ solution for the same problem (measured speedup: 5.252147377392458x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

void init(){

    cin.tie(0);

    cin.sync_with_stdio(0);

}

int fr[205];

int main()

{   init();

    int n;

    cin>>n;

    map<string ,int>mp;

    for(int i=0;i<n;i++){

        string s;

        cin>>s;

        mp[s]++;}

        cout<<"AC x "<<mp["AC"]<<endl<<"WA x "<<mp["WA"]<<endl<<"TLE x "<<mp["TLE"]<<endl<<"RE x "<<mp["RE"];



}


// Faster C++ solution for the same problem (measured speedup: 5.338852722137241x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <iostream>

#include<bits/stdc++.h>

using namespace std;



int main() {

int n;

cin>>n;

int AC=0,WA=0,TLE=0,RE=0;

while(n--){

	string s;

	cin>>s;

	if(s=="AC")AC++;

	else if(s=="WA")WA++;

	else if(s=="TLE")TLE++;

	else

	RE++;

}cout<<"AC x "<<AC<<endl;

cout<<"WA x "<<WA<<endl;

cout<<"TLE x "<<TLE<<endl;

cout<<"RE x "<<RE<<endl;



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.251592716745243x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;cin>>n;

    string s;

    string arr[4]={"AC" , "WA" ,"TLE" ,"RE" };

    int intarr[4]={};

    for(int i=0;i<n;i++)

    {

        cin>>s;

        if(s=="AC")

            intarr[0]++;

        else if(s=="WA")

            intarr[1]++;

        else if(s=="TLE")

            intarr[2]++;

        else intarr[3]++;

    }

    for(int i=0;i<4;i++)

    {

        cout<<arr[i]<<" x "<<intarr[i]<<endl;

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.252065241233898x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
// 2nd.cpp : This file contains the 'main' function. Program execution begins and ends there.

//



#include <iostream>

#include<string>

#include<vector>

using namespace std;



int main()

{

	int n;

	cin >> n;

	int a = 0;

	int b = 0;

	int c = 0;

	int d = 0;



	vector<string> ans(n);

	for (int i = 0; i < n; i++)

	{

		cin >> ans[i];

		if (ans[i] == "AC")

			a++;

		else if (ans[i] == "WA")

			b++;

		else if (ans[i] == "TLE")

			c++;

		else

			d++;





	}

	cout << "AC" << " " << "x" << " " << a << endl;

	cout << "WA" << " " << "x" << " " << b << endl;

	cout << "TLE" << " " << "x " << c << endl;

	cout << "RE" << " x " << d << endl;





	return 0;



}



// Run program: Ctrl + F5 or Debug > Start Without Debugging menu

// Debug program: F5 or Debug > Start Debugging menu



// Tips for Getting Started: 

//   1. Use the Solution Explorer window to add/manage files

//   2. Use the Team Explorer window to connect to source control

//   3. Use the Output window to see build output and other messages

//   4. Use the Error List window to view errors

//   5. Go to Project > Add New Item to create new code files, or Project > Add Existing Item to add existing code files to the project

//   6. In the future, to open this project again, go to File > Open > Project and select the .sln file


// Faster C++ solution for the same problem (measured speedup: 5.261656328784278x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <iostream>



using namespace std;

int main ()

{

    /*/

    int i = 0, j = 1;

    cout << i;

    int *ptr = &i;

    cout << i;



    i = 2;

    cout << i;

    *ptr = 3;

    ptr = &j;

    j = i;

    cout << i;

    *ptr = 4;

      /*/



    /*/

    int n , ctn=0 ,arr;

    cin >> n ;

    for (int i = 1; i <=n ; ++i) {

        cin >> arr;



        if ((i%2!=0)&&(arr%2!=0))ctn++;



    }

    cout<<ctn <<endl;



    cout << "Hello, world!"<<endl;

    cout << "Greetings from Illinois!" <<endl;

    /*/



    int n ,ac=0,wa=0,tle=0,re=0 ;

    string inp;

    cin >> n ;

    for (int i = 0; i < n ; ++i)

    {

        cin >> inp;

        if (inp =="AC") ac++;

        if (inp=="WA") wa++;

        if (inp=="TLE") tle++;

        if (inp=="RE") re++;



    }

    cout <<"AC x "<< ac <<endl;

    cout <<"WA x "<< wa <<endl;

    cout <<"TLE x "<< tle <<endl;

    cout <<"RE x "<< re <<endl;





    

}

// Faster C++ solution for the same problem (measured speedup: 5.253458173852292x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



map<string, int> mp;

int main() {

    int n;

    cin >> n;

    for (int i = 1; i <= n; i++) {

        string s;

        cin >> s;

        mp[s]++;

    }

    cout << "AC x " << mp["AC"] << endl;

    cout << "WA x " << mp["WA"] << endl;

    cout << "TLE x " << mp["TLE"] << endl;

    cout << "RE x " << mp["RE"] << endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.2721243406157505x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

char st[10];

int n;

map <string,int> mp;

int main(){

//	freopen("x.txt","r",stdin);

	scanf("%d\n",&n);

	for(int i=1;i<=n;++i){

		cin>>st;

//		cout<<st<<endl;

		mp[st]++;

	}

	cout<<"AC x "<<mp["AC"]<<endl;

	cout<<"WA x "<<mp["WA"]<<endl;

	cout<<"TLE x "<<mp["TLE"]<<endl;

	cout<<"RE x "<<mp["RE"]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.291247570943899x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

const int MAXN = 1e5 + 5;



int a, b, c, d;

char str[500];



int main(int argc, char* argv[]) {

    int T;

    scanf("%d", &T);

    while (T--) {

        scanf("%s", str);

        if (str[0] == 'A')

            a++;

        else if (str[0] == 'W')

            b++;

        else if (str[0] == 'T')

            c++;

        else if (str[0] == 'R')

            d++;

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", a, b, c, d);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.133309884265321x):
#include <stdio.h>



int main(){



    int AC=0;

    int TLE=0;

    int WA=0;

    int RE=0;

    int brpKali;

    scanf("%d", &brpKali);

    char inputan[brpKali][100];

    for(int i=0;i<brpKali;i++){

        scanf("%s", inputan[i]);    //AC, TLE, AC, AC, WA ,TLE



    }

    for(int j=0;j<brpKali;j++){

        if(inputan[j][0]=='A'){

            AC++;

        }

        else if(inputan[j][0]=='W'){

            WA++;

        }

        else if(inputan[j][0]=='T'){

            TLE++;

        }

        else if(inputan[j][0]=='R'){

            RE++;

        }



    }

printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", AC, WA, TLE, RE);









    return 0;

}