stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int main(){

    int c1=0,c2=0,c3=0,c4=0,x;

    string s;

    cin>>x;

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

        cin>>s;

        if(s=="AC"){

            c1++;

        }

        else if(s=="WA")

            c2++;

        else if (s=="TLE")

            c3++;

        else if(s=="RE")

            c4++;

        

    }



    cout<<"AC x "<<c1<<endl;

    cout<<"WA x "<<c2<<endl;

    cout<<"TLE x "<<c3<<endl;

    cout<<"RE x "<<c4<<endl;







    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.25149264401841x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

using namespace std;

int main()

{

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

	string test;

	cin>>n;

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

	{

		cin>>test;

		if (test=="AC")

			AC++;

		else if (test=="TLE")

			TLE++;

		else if (test=="WA")

			WA++;

		else if (test=="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: 5.2511482248646475x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

typedef long long ll;

#define dbg(x) cout<<#x<<" = "<<x<<endl

#define dbg2(x,y) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<endl

#define dbg3(x,y,z) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<", "<<#z<<" = "<<z<<endl

#define dbg4(x,y,z,q) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<", "<<#z<<" = "<<z<<", "<<#q<<" = "<<q<<endl

#define dbg_array(x,sz) for (int i = 0; i < sz; ++i) cout << x[i] << " \n"[i==sz-1]

using namespace std;



const ll N = 100 + 9, M = 400 - 9, Max_ans = 1LL * (1e16);

int vis[N];



int main() {

	ios_base::sync_with_stdio(0), cin.tie(0);

#ifdef ECLIPSE

	freopen("input.in", "rt", stdin);

	//freopen("output.out", "wt", stdout);

#endif

	int n;

	cin >> n;

	string s;

	while (n--) {

		cin >> s;

		++vis[s[0] - 'A'];

	}

	printf("AC x %d\n", vis['A' - 'A']);

	printf("WA x %d\n", vis['W' - 'A']);

	printf("TLE x %d\n", vis['T' - 'A']);

	printf("RE x %d\n", vis['R' - 'A']);

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.259477081876113x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

#include <map>

#include <string>

using namespace std;



int main()

{

	map<string, int> test;

	int count = 0;

	string verdict;

	cin >> count;

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

	{

		test["AC"] = 0;

		

		test["WA"] = 0;

		

		test["TLE"] = 0;



		test["RE"] = 0;

	}



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

	{

		cin >> verdict;

		

		if (verdict == "AC")

			test["AC"]++;

		else if (verdict == "WA")

			test["WA"]++;

		else if (verdict == "TLE")

			test["TLE"]++;

		else if (verdict == "RE")

			test["RE"]++;

	}

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

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

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

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.294815549776055x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

using namespace std;

#define ll long long



bool cmp(const pair<string, int>&i, const pair<string, int>&j)

{

    return i.second > j.second;

}







int main() {

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

            int n ; cin >> n ;  string s ;

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

                  cin >> s;

                   if ( s == "AC")

                       AC++;

                   else if ( s == "TLE")

                       TLE++;

                   else if ( s == "WA")

                       WA++;

                   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: 5.250683041631725x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

#include<string>

using namespace std;

int main(){

	int t, c0 = 0, c1 = 0, c2 = 0, c3  = 0;

	cin>>t;

	string test;

	while(t--){

		cin>>test;

		if(test == "AC")

		   c0++;

		else if(test == "WA")

		   c1++;

		else if(test == "TLE")

		   c2++;

		else if(test == "RE")

		   c3++;

	}

	cout<<"AC"<<" x "<<c0<<endl;

	cout<<"WA"<<" x "<<c1<<endl;

	cout<<"TLE"<<" x "<<c2<<endl;

	cout<<"RE"<<" x "<<c3<<endl;

	

	

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.25187184322697x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

#include <string>



using namespace std;



int main()

{

    int N;

    cin>>N;



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

    string s;



    while(N--)

    {

        cin>>s;



        if(s.compare("AC")==0)

            AC++;



        if(s.compare("WA")==0)

            WA++;



        if(s.compare("TLE")==0)

            TLE++;



        if(s.compare( "RE")==0)

            RE++;





    }



    cout<<"AC x "<<AC<<"\n";

    cout<<"WA x "<<WA<<"\n";

    cout<<"TLE x "<<TLE<<"\n";

    cout<<"RE x "<<RE<<"\n";

}


// Faster C++ solution for the same problem (measured speedup: 5.24706060231326x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

#include<cstdio>

#include<map>

using namespace std;

int main(void)

{

    ios::sync_with_stdio(false);

    map<string,int> mp;

    int n;

    char ch[10];

    while(cin>>n)

    {

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

        cin>>ch,mp[ch]++;

        printf("AC x %d\n",mp["AC"]);

        printf("WA x %d\n",mp["WA"]);

        printf("TLE x %d\n",mp["TLE"]);    

        printf("RE x %d\n",mp["RE"]);

    }

    return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.325584787976921x):
#include <stdio.h>

#include <string.h> // strmcp(s1, s2)

// contoh soal menggunakan switch case

// 1. scan semua input

// 2. cari jumlah ac, wa, tle, re

//strcmp membandingkan apakah str 1 dgn str2

int main(){

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

	scanf("%d", &N);

	

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

		char str[10];

		scanf("%s", str);

		

		if(strcmp (str, "AC") == 0) {

		AC++;

		} else if(strcmp (str, "WA") == 0) {

		WA++;

	}else if(strcmp (str, "TLE") == 0) {

		TLE++;

	}else if(strcmp (str, "RE") == 0) {

		RE++;

	}

	}

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

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

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

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

	return 0;

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

#include<string>

#include<stdio.h>

#include<stdlib.h>

#include<set>

#include<map>

#include<vector>

#include<queue>

using namespace std;

#define mp(a,b) make_pair(a,b)

#define pb push_back



int main(){

    int n;

    cin>>n;

    int a[5];

    for (int i=1;i<=4;i++){

        a[i] = 0;

    }

    string sss;

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

        cin>>sss;

        if (sss == "AC"){

            a[1]++;

        }

        if (sss == "WA"){

            a[2]++;

        }

        if (sss == "TLE"){

            a[3]++;

        }

        if (sss == "RE"){

            a[4]++;

        }

    }

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

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

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

    cout<<"RE x "<<a[4]<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.256992239547174x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

}
// Slower C++ solution:
// In the name of God



#include <bits/stdc++.h>



using namespace std;



int main() {

	ios_base::sync_with_stdio(false);

	cin.tie(0);



	map<string, int> mp;



	int n;

	cin >> n;

	while(n--) {

		string s;

		cin >> s;

		mp[s]++;

	}



	cout << "AC x " << mp["AC"] << "\n";

	cout << "WA x " << mp["WA"] << "\n";

	cout << "TLE x " << mp["TLE"] << "\n";

	cout << "RE x " << mp["RE"] << "\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.346131086482734x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;

using namespace std;



typedef long long ll;

const int N = 2e5 + 10, OO = 0x3f3f3f3f, mod = 1e9 + 7;



#define PI acos(-1)

#define clr(arr, val) memset(arr, val, sizeof(arr))

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

#define rloop(i, n) for (int i = int(n) - 1; i >= 0; i--)

#define xloop(i, a, b) for (int i = int(a); i <= int(b); i++)

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

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

//int dx[] = {1, 0, -1, 0};

//int dy[] = {0, 1, 0, -1}; //4 Direction

int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};

int dy[] = {0, 1, 1, 1, 0, -1, -1, -1}; //8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction



int n;

int main()

{

  //ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);



  cin >> n;

  string s;

  map < string , int > cnt;

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

    cin >> s;

    cnt[s]++;

  }



  cout << "AC x " << cnt["AC"] << "\n";

  cout << "WA x " << cnt["WA"] << "\n";

  cout << "TLE x " << cnt["TLE"] << "\n";

  cout << "RE x " << cnt["RE"] << "\n";





  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.274091149061417x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>

#include <unordered_map>

#define endl '\n';

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

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

#define Ceil(x,y)             ((x+y-1)/y)

#define RT(x)           return cout<<x,0;

#define mem(x,y)    memset(x,y,sizeof(x))

#define vi                    vector<int>

#define pii                 pair<int,int>

#define fx(n)           fixed<<setprecision(n)

#define watch(x) cout<<(#x)<<" = "<<x<<endl

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

typedef long long ll;

typedef unsigned long long ull;

typedef double dl;

const double PI = acos(-1), EPS = 1e-7;

const int OO = 0x3f3f3f3f, N = 1e7 + 5, MOD = 1e9 + 7;

using namespace std;

void fast() { std::ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL); }

int main()

{

	fast();

	int n;

	cin>>n;

	map<string,int>mp;

	string s;

	while(n--)

    {

        cin>>s;

        mp[s]++;

    }

    cout<<"AC x "<<mp["AC"]<<"\n";

    cout<<"WA x "<<mp["WA"]<<"\n";

    cout<<"TLE x "<<mp["TLE"]<<"\n";

    cout<<"RE x "<<mp["RE"]<<"\n";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.383803994902428x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

using namespace std;

int main() {

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

    cin >> n;

    string s;

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

        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: 5.256238529981107x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

using namespace std;



int main()

{

    int n,nAC=0,nWA=0,nTLE=0,nRE=0,i;

    string in;

    cin>>n;

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

    {

        cin>>in;

        if(in=="AC")

            nAC++;

        else if(in=="WA")

            nWA++;

        else if(in=="TLE")

            nTLE++;

        else

            nRE++;

    }

    cout<<"AC x "<<nAC<<endl;

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.256269954690404x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

#include<cstdlib>



/*If you are going to try,keep going until the end ,or don't even try to start*/



/* Remember, Hope is a good thing... May be the best of things... And No Good thing ever Dies !!!*/



using namespace std;



#define ll long long

#define nada "\n"

#define st string

#define S  second

#define F  first

#define pb push_back

#define MP make_pair

#define isnttriangle(a,b,c)  (a+b<c||a+c<b||b+c<a)

#define todegree(rad) rad*(180.0/pi)

#define rrp(i, a, b)  for(ll i = a; i >= b; i--)

#define repp(i,a,b)   for (int i = int(a); i <=int(b); i++)

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

#define rep1(i,n)     for (int i = 1; i <=n; i++)

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

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

#define fx(n)         cout<<fixed<<setprecision(n);

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

#define Test          ll t;cin>>t;while(t--)

#define mod           1000000009LL





const double ME = 2.71828;

const int    N  = 3e5 + 20;

const int   inf = 0x3f3f3f3f;

const double pi = 3.141592653;









void LOSER()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



void file()

{

    freopen("records.in", "r", stdin);

    //freopen("text.out", "w", stdout);

}



ll gcd(ll a, ll b)

{

    //Eculidean Algorthim

    return (!b) ? a : gcd(b, a % b);

    //complixty=log(ab);

}



ll lcm(ll a,ll b)

{

    return a / gcd(a, b) * b;

}



ll Fib_recur(ll n)

{

    vector<ll>v(n+5,0);

    //BaseCase

    v[1]=0;

    v[2]=1;

    repp(i,3,n)v[i]=v[i-1]+v[i-2];

    return v[n];

}



bool isprime(int n)

{

    for(int i = 2; i <= sqrt(n); i++)

    {

        if(n%i == 0)

            return false;

    }

    return true;

}



void Solve()

{



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

  st s;

  cin>>n;

  while(n--)

  {

      cin>>s;

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

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

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

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

  }

  cout<<"AC"<<" x "<<AC<<nada;

  cout<<"WA"<<" x "<<WA<<nada;

  cout<<"TLE"<<" x "<<TLE<<nada;

  cout<<"RE"<<" x "<<RE<<nada;

}











int main()

{

    // fx(7)

    //file();

    LOSER();

    //Test



    Solve();















}

// Faster C++ solution for the same problem (measured speedup: 5.32111943516744x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

using namespace std;

int main()

{

	int N, counter_ac = 0, counter_wa = 0, counter_tle = 0, counter_re = 0;

	cin >> N;

	string s;

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

	{

		cin >> s;

		if (s == "AC")

		{

			counter_ac++;

		}

		else if (s == "WA")

		{

			counter_wa++;

		}

		else if (s == "TLE")

		{

			counter_tle++;

		}

		else if (s == "RE")

		{

			counter_re++;

		}

	}

	cout << "AC x " << counter_ac << endl;

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.255961992539291x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

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

#include <string>

#include <algorithm>

#include <cmath>

using namespace std;

int main()

{  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

   // your code goes here

    int t;long long c0=0,c1=0,c2=0,c3=0;

    string s;

    cin>>t;

    for (int i=1;i<=t;i++)

    {

        cin>>s;

        if (s=="AC")c0++;

        else if (s=="WA")c1++;

        else if (s=="TLE")c2++;

        else if (s=="RE")c3++;

    }

    cout<<"AC x "<<c0<<endl;

    cout<<"WA x "<<c1<<endl;

    cout<<"TLE x "<<c2<<endl;

    cout<<"RE x "<<c3<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.320527200261454x):
#include <stdio.h>





int main(){

	int N, dec[4]= {0};

	scanf("%d", &N);

	

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

		char S[5];

		scanf("%s", S);

		switch(S[0]){

			case 'A':

				dec[0]++;

				break;

			case 'W':

				dec[1]++;

				break;

			case 'T':

				dec[2]++;

				break;

			case 'R':

				dec[3]++;

				break;

		}

	}

	printf("AC x %d\n", dec[0]);

	printf("WA x %d\n", dec[1]);

	printf("TLE x %d\n", dec[2]);

	printf("RE x %d\n", dec[3]);

	

	

	return 0;

}