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

using namespace std;

int n;

char s[10];

int cnta,cntw,cntt,cntr;

int main()

{

	cnta = cntw = cntt = cntr = 0;

	scanf("%d", &n);

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

	{

		scanf("%s",s);

		if(s[0] == 'A') cnta ++;

		if(s[0] == 'W') cntw ++;

		if(s[0] == 'T') cntt ++;

		if(s[0] == 'R') cntr ++; 

	}

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

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

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

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.133372332079963x):
#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 a[4] = {0, 0, 0, 0};

    int i;

    cin >> i;

    while (i--) {

        string s;

        cin >> s;

        if (s == "AC")a[0]++;

        if (s == "WA")a[1]++;

        if (s == "RE")a[2]++;

        if (s == "TLE")a[3]++;

    }

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

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.253500709126673x):
#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:
//	Id:lys2002

//	Time:2020/8/7 15:24

#include<bits/stdc++.h>

using namespace std;



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

#define per(i,a,n) for (int i=n-1;i>=a;i--)



typedef double db;

typedef long long ll;



const ll md=1000000007;



inline void Fio(){

	freopen("1.in","r",stdin);

	freopen("1.out","w",stdout);

}

map <string,int> mp;

int main(){

    cin.tie(0);

    ios::sync_with_stdio(false);

    int n;

    cin>>n;

    string str;

    rep(i,0,n){

        cin>>str;

        if(!mp.count(str))

                mp[str]=0;

        mp[str]++;

    }

    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: 5.3401315458961465x):
#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.h>

#include<algorithm>



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++;

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

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

    if(s=="RE")  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: 5.252021590685128x):
#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 <bits/stdc++.h>



using namespace std;



int main()

{

    int N, acCnt = 0, waCnt = 0, tleCnt = 0, reCnt = 0;

    cin >> N;

    string AC = "AC", WA = "WA", TLE = "TLE", RE = "RE", x;

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

    {

        cin >> x;

        if(x == AC)

            acCnt++;

        else if(x == WA)

            waCnt++;

        else if(x == TLE)

            tleCnt++;

        else if(x == RE)

            reCnt++;

    }

    cout << "AC x " << acCnt << endl;

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

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

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.250618795141613x):
#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;

struct test {

string T;

int n;



};

int main(){

    test test_cases[4];

    test_cases[0].T = "AC", test_cases[1].T = "WA", test_cases[2].T = "TLE", test_cases[3].T = "RE" ;

    test_cases[0].n = 0, test_cases[1].n = 0, test_cases[2].n = 0, test_cases[3].n = 0;

    int m;

    cin>>m;

    string s;

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

        cin>>s;

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

            if(s == test_cases[j].T){

                test_cases[j].n++;

            }

        }

    }

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

        cout<<test_cases[i].T<<" x "<<test_cases[i].n<<endl;

    }

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.240321000583532x):
#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,C0=0,C1=0,C2=0,C3=0;

cin>>N;

string s[N];



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

{ cin>>s[i];

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

    C0++;

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

        C1++;

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

        C2++;

    else if(s[i]=="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: 5.252306593911944x):
#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<vector>

#include <cstdlib>

#include<cmath>

#include<string>

#include <iomanip>

#include <sstream>



using namespace std;







int main (){



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

    int a[4]={0,0,0,0};







    int n,i,j,cnt=0;

    cin>>n;



    string words[n];



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



        cin>>words[i];



    }



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



        for(j=0;j<n;j++){



            if( s[i] == words[j] ){



                cnt++;

            }



        }



        a[i]=cnt;

        cnt=0;



    }





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



    cout<<s[i]<<" "<<"x"<<" "<<a[i];

    cout<<endl;



}













}

// Faster C++ solution for the same problem (measured speedup: 5.242864292090629x):
#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 x,a=0,w=0,t=0,r=0;

cin>>x;

string arr[x];

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

    cin>>arr[i];

    if(arr[i]=="AC"){a++;}

    else if(arr[i]=="WA"){w++;}

    else if(arr[i]=="TLE"){t++;}

    else if(arr[i]=="RE"){r++;}

}

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



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



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



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





    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.254110809254586x):
#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;

typedef long long int ll;

const int N=3e5+9;





int main()

{



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    int n;

    string s;

    map<string,int>cnt;

    cin>>n;

    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: 5.333055253913722x):
#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 <map>



using namespace std;



map<string, int>mymap;

int main() {

	int N;

	cin >> N;

	mymap["AC"] = 0;

	mymap["WA"] = 0;

	mymap["TLE"] = 0;

	mymap["RE"] = 0;

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

		string s;

		cin >> s;

		mymap[s]++;

	}

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

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

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

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.276750209646441x):
#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;

    string value;

    int c[4] = {0,0,0,0};

    cin>> n;

    while(n){

        cin>> value;

        if(value == "AC")

            c[0]+= 1;

        else if(value == "WA")

            c[1]+= 1;

        else if(value == "TLE")

            c[2]+= 1;

        else

            c[3]+= 1;

        n--;

    }

    cout<< "AC x "<< c[0] <<"\nWA x "<< c[1] <<"\nTLE x "<< c[2] << "\nRE x "<< c[3];

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.241545548346209x):
#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 <bits/stdc++.h>

#include<sstream>

using namespace std;

const int N = 1e5 + 7, M = 1e7, OO = 0x3f3f3f3f;

#define endl '\n'

#define  ll long long

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

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

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

#define p(a) printf("%d ",a)

#define S(n) scanf("%d",&n)

int main () {

    map<string,int>mp;

    mp["AC"]=0;

    mp["WA"]=0;

    mp["TLE"]=0;

    mp["RE"]=0;

    int n;S(n);

    string s;

    loop(i,n) {

        cin >> s;

        mp[s]++;

    }

    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: 5.247734948931286x):
#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<vector>

#include<map>

#include<set>

#include<string>

#include<algorithm>

#include<math.h>

#include<bits/stdc++.h>

#include<cmath>

#include<iomanip>

using namespace std;



const double PI  =3.141592653589793238463;

#define pb push_back

#define pf push_front

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

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

typedef unsigned long long ull;

typedef long long ll;



typedef vector<int> vi;

typedef pair<int,int> pi;

//2 7 3 4 9



void printVec(vi v1)

{

    for(int i=0; i<v1.size(); i++)

        cout<<v1[i]<<" ";

    cout<<endl;

}

int main()

{

    ios::sync_with_stdio(false);

    cout.tie(NULL);

    cin.tie(NULL);





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



long long t;cin>>t;

string s;

while(t--){

        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: 5.312440342544895x):
#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<bits/stdc++.h>

#define input ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define ll long long

#define ull unsigned long long

#define ii pair<int,int>

#define dl endl

#define ft first

#define sc second

using namespace std;

int n, a[4];

string s;

int main()

{

	//freopen("", "r", stdin);

	//freopen("", "w", stdout);

	input;

	cin >> n;

	while (n--)

	{

		cin >> s;

		if (s == "AC")a[0]++;

		else if (s == "WA")a[1]++;

		else if (s == "TLE")a[2]++;

		else a[3]++;

	}

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

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.278224497524819x):
#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;

    cin >> N;

    string S[100000];

    int A = 0, W = 0, T = 0, R = 0;

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

        cin >> S[i];

    }

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

        if (S[i] == "AC") {

            A ++;

        }

    }

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

        if (S[i] == "WA") {

            W ++;

        }

    }

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

        if (S[i] == "TLE") {

            T ++;

        }

    }

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

        if (S[i] == "RE") {

            R ++;

        }

    }

    cout << "AC x " << A<<endl;

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 22.925525758495027x):
#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 <set>

#include <string>



using namespace std;



int main() {

    int n; cin >> n;

    string temp;

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

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

        cin >> temp;

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

        else if(temp == "WA") wa++;

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

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

    }

    cout << "AC x " << ac << "\n"

        << "WA x " << wa << "\n"

        << "TLE x " << tle << "\n"

        << "RE x " << re << "\n";

}

// Faster C++ solution for the same problem (measured speedup: 5.245850063183767x):
#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;

}