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

using namespace std;

int main(){

    int tc;cin>>tc;

    int acCount,waCount,tleCount,reCount;

    acCount = waCount = tleCount = reCount = 0;

    while(tc > 0){

        char arr[4];

        cin>>arr;

        if(arr[0] == 'A' && arr[1] == 'C')

            acCount++;

        else if(arr[0] == 'W' && arr[1] == 'A')

            waCount++;

        else if(arr[0] == 'R' && arr[1] == 'E')

            reCount++;

        else

            tleCount++;

        tc--;

    }

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

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

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

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.192052462886037x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;



int main()

{

    int n;

    cin>>n;

    string s;

    int arr[4]={0};

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

        cin>>s;

        if (s=="AC")

            arr[0]++;

        else if (s=="WA")

            arr[1]++;

        else if (s=="TLE")

            arr[2]++;

        else

            arr[3]++;

    }

    cout<<"AC x "<<arr[0]<<"\n";

    cout<<"WA x "<<arr[1]<<"\n";

    cout<<"TLE x "<<arr[2]<<"\n";

    cout<<"RE x "<<arr[3];



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.234788000107996x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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:
#define ll long long int



#include <bits/stdc++.h>



using namespace std;



int main() {



 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;

    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: 5.26473517852115x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;

    

    string Si[N];

    for (int i{}; i<N; i++) 

        {

            cin >> Si[i];

        }

    int C[4] {0,0,0,0};

    for (auto i : Si) 

        {

            if (i == "AC")

                C[0] += 1;

            if (i == "WA")

                C[1] += 1;

            if (i == "TLE")

                C[2] += 1;

            if (i == "RE")

                C[3] += 1;

        }

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

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

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

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

        

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.245448260301077x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 <vector>

#include <algorithm>

#import<iostream>

#include <bits/stdc++.h>

using namespace std;



int main()

{

    string arr[5]= {"AC","WA","TLE","RE"};

    int C0=0,C1=0,C2=0,C3=0;

    int N;

    cin>>N;

    string arr_2[N];

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

    {

        cin>>arr_2[i];



        if(arr_2[i]=="AC")

            C0+=1;

        else if(arr_2[i]=="WA")

            C1+=1;

        else if(arr_2[i]=="TLE")

            C2+=1;

        else if(arr_2[i]=="RE")

            C3+=1;

    }



    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: 5.238726477133787x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 <algorithm>

using namespace std;



int main()

{

   int n,c=0,w=0,t=0,r=0;

   cin>>n;

   string a[n];

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

    cin>>a[i];

   }

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

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

        c++;

    }

    else if(a[i]=="WA"){

        w++;

    }

    else if(a[i]=="TLE"){

        t++;

    }

    else if(a[i]=="RE"){

        r++;

    }

   }

   cout<<"AC x "<<c<<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: 5.198545236800388x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 arr_size, AC_count = 0, WA_count = 0, TLE_count = 0, RE_count = 0;

    string array[100000], in;

    cin >> arr_size;

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

        cin >> array[i];

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

    {

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

            AC_count++;

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

            WA_count++;

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

            TLE_count++;

        else if (array[i] == "RE")

            RE_count++;

    }

    cout << "AC x " << AC_count << endl << "WA x " << WA_count << endl << "TLE x " << TLE_count << endl << "RE x " << RE_count << endl;

}




// Faster C++ solution for the same problem (measured speedup: 22.856942017749905x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

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

int T;

cin>>T;

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

{

    string s;

    cin>>s;

 if(s=="AC")

        AC++;

 else if(s=="WA")

    WA++;

 else if(s=="TLE")

    TLE++;

 else if(s=="RE")

    RE++;



}

cout<<"AC x "<<AC<<endl<<"WA x "<<WA<<endl<<"TLE x "<<TLE<<endl<<"RE x "<<RE<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.240758632925648x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 <cstring>



using namespace std;



int main()

{

    int n, c0, c1, c2, c3, sVal;

    c0 = c1 = c2 = c3 = 0;

    cin >> n;

    char c [4];

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

        cin >> c;

        sVal = strcmp(c, "RE");

        if ( sVal ==0) c3++;

        else if(sVal < 0) c0++;

        else {

            if (strcmp(c, "WA")==0) c1++;

            else c2++;

        }

    }

    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: 5.195308251275702x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 <cstring>

using namespace std;

int main()

{

    int N, t = 0, y = 0, z = 0, c = 0, i;

    cin >> N;

    string a[N];

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

        cin >> a[i];

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

            t = t + 1;

        if (a[i] == "WA")

            y = y + 1;

        if (a[i] == "TLE")

            z = z + 1;

        if (a[i] == "RE")

            c = c + 1;

    }

    cout << "AC x " << t << endl;

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.242838519067069x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;



int main() {

   int N;

   cin >> N;

   map<string, int> ans;

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

       string s;

       cin >> s;

       ans[s]++;

   }

   for (string s: {"AC", "WA", "TLE", "RE"}) {

       cout << s << " x " << ans[s] << endl;

   }

}

// Faster C++ solution for the same problem (measured speedup: 5.305032128792876x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;

    int a=0,w=0,t=0,r=0;

    cin >>n;

    string arr[n];

    for(int i=0;i<n;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: 5.242177526044379x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 x, q = 0, w = 0, e = 0, r = 0;

	string a = "AC", b = "WA", c = "RE", t = "TLE",m;

	cin >> x;

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

	{

		cin >> m;

		if (m==a)

		{

			q++;

		}

		else if (m == b) {

			w++;

		}

		else if (m == t) {

			e++;

		}

		else if (m == c) {

			r++;

	}

	}

	cout << "AC x " << q << endl;

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

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

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

}


// Faster C++ solution for the same problem (measured speedup: 5.23163394003911x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;

int main()

{

	map<string, int> m;

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

	int n;

	cin>>n;

	while (n--)

	{

		string s;

		cin>>s;

		m[s]++;

	}

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

	{

		cout<<ar[i]<<" x "<<m[ar[i]]<<"\n";

	}

}

// Faster C++ solution for the same problem (measured speedup: 5.2533329283485495x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;

using namespace std;

int main()

{

	ios_base::sync_with_stdio(0);

	cin.tie(0);cout.tie(0);

	ll n,a=0,b=0,c=0,d=0;

	string s;

	cin >> n;

	while(n--){

		cin >> s;

		if(s=="AC"){

			a++;

		}

		else if(s=="TLE"){

			b++;

		}

		else if(s=="WA"){

			c++;

		}

		else if(s=="RE"){

			d++;

		}

	}

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

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.304256872206087x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;



int N;

int arr[5];

string str;



int main() {

	

	ios_base::sync_with_stdio(false);

	cin >> N;

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

		cin >> str;

		if(str == "AC") arr[1]++;

		else if(str == "WA") arr[2]++;

		else if(str == "TLE") arr[3]++;

		else if(str == "RE") arr[4]++;

	}

	cout << "AC x " << arr[1] << "\n";

	cout << "WA x " << arr[2] << "\n";

	cout << "TLE x " << arr[3] << "\n";

	cout << "RE x " << arr[4] << "\n";

}

// Faster C++ solution for the same problem (measured speedup: 5.260965927280161x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 <cmath>

#include <string>

using namespace std;

int main()

{

	string arr[100000];

	int countAC = 0, countTLE = 0, countWA = 0 , countRE=0;

	int num;

	cin >> num;

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

	{

		cin >> arr[i];

	}

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

	{

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

			countAC++;

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

			countTLE++;

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

			countWA++;

		else if (arr[i] == "RE")

			countRE++;



	}

	cout << "\nAC"<<" "<<"x"<<" " << countAC;

	cout << "\nWA" << " " << "x" <<" "<< countWA;

	cout << "\nTLE"<<" "<<"x" << " " << countTLE;

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 22.80773540222703x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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 = 0 , 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 << "\n"

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

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

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.235701626496033x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

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

		AC++;

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

		WA++;

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

		TLE++;

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

 		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;

int main()

{

    map<string,int> m;

    int n;

    cin>>n;

    string x;

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

    {

        cin>>x;

        m[x]++;

    }

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

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.270223071552656x):
#include <stdio.h>



int main() {

    int cc;

    scanf("%d", &cc);

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

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

   char k[100];

   scanf("%s", k);

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

        AC++;

     }

     if(k[0] == 'T'){

       TLE++;

     }

     if(k[0]== 'W'){

         WA++;

     }

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

        RE++;

     }

   



    }



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



     return 0;

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

using namespace std;

int main()

{

 int n,a=0,r=0,t=0,w=0;

 cin>>n;

 string s;

 while(n--)

 {

     cin>>s;

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

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

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

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

 }

 cout<<"AC x "<<a<<"\nWA x "<<w<<"\nTLE x "<<t<<"\nRE x "<<r;



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.242923095790117x):
#include <stdio.h>



int main() {

    int cc;

    scanf("%d", &cc);

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

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

   char k[100];

   scanf("%s", k);

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

        AC++;

     }

     if(k[0] == 'T'){

       TLE++;

     }

     if(k[0]== 'W'){

         WA++;

     }

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

        RE++;

     }

   



    }



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



     return 0;

}