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

#include<algorithm>

#include<math.h>

 #define ll long long

using namespace std;

int main()

{

	int co1=0, co2=0, co3=0, co4=0;

    string s; int t; cin >> t;

	while (t--)

	{

		cin >> s;

		if (s == "AC")

			co1++;

		else if (s == "WA")

			co2++;

		else if (s == "TLE")

			co3++;

		else {

			co4++;

		}

		

	}

	cout << "AC x " << co1 << endl;

	cout << "WA x " << co2 << endl;

	cout << "TLE x " << co3 << endl;

	cout << "RE x " << co4 << endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.252789542801979x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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 endl "\n"

#define goat() ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);srand(time(NULL))

typedef long long ll;

int a[1005]; int b[1005];

int main() {

	goat();

	int n, m, i;

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

	cin >> n;

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

		string s;

		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;

	

}

// Faster C++ solution for the same problem (measured speedup: 5.319396490495697x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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=0,c_AC=0,c_WA=0,c_TLE=0,c_RE=0;

cin>>n;

while (n>0)

{

    string s;

    cin>>s;

    if ( s[0]=='A'){c_AC++;}

    if ( s[0]=='W'){c_WA++;}

    if ( s[0]=='T'){c_TLE++;}

    if ( s[0]=='R'){c_RE++;}

    n--;

}

cout<<"AC x "<<c_AC<<endl;

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.244067963669992x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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 test;   cin >> test;

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



    string stat;

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



        cin >> stat;

        if (stat == "AC"){AC++;}

        if (stat == "WA"){WA++;}

        if (stat == "TLE"){TLE++;}

        if (stat == "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: 5.253183803982617x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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()

{

    long long n;

cin>>n;

    string str;

    long long a=0,b=0,c=0,d=0;

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

{

    cin>>str;

    transform(str.begin(), str.end(),str.begin(), ::toupper);



    if(str=="AC")

        a++;

    else if(str=="WA")

        b++;

    else if(str=="TLE")

    c++;

    else if(str=="RE")

        d++;

        else

            continue;

}

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

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

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

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



}

// Faster C++ solution for the same problem (measured speedup: 5.251901005655909x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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 s;

    cin>>n;

    while(n--){

        cin>>s;

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

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

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

        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: 5.253559704961217x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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()

{

    long long int n;

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



    cin>>n;



    string S[n];

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

        cin>>S[i];

    }



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

        if(S[j] == "AC"){AC++;}

        else if(S[j] == "WA"){WA++;}

        else if(S[j] == "TLE"){TLE++;}

        else if(S[j] == "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: 5.252482734162634x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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>

#define ll long long

using namespace std;

int main(){

	ll n;

	map<string,ll>m;

	cin >> n;

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

		string s;

		cin >> s;

		m[s]++;

	}

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

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.270005131193307x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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;





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: 5.245630030332986x):
#include <stdio.h>



int main() {

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

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

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

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			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>

#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: 5.2530756195680315x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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 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: 5.250754533966588x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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 <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: 5.201860539709162x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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;

/*  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: 5.266540814394306x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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;

//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: 5.243140600303746x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

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

    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: 5.256731389772345x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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 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: 5.272974158419285x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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;

    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: 5.251086876712435x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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;

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: 5.337774714127272x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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 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: 5.250532328060935x):
#include <stdio.h>

#include <string.h> // strcmp(s1, s2) -> membandingkan apakah s1 sama dengan s2



// 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[10];

    scanf("%s", str);



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

      AC++;

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

      WA++;

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

      TLE++;

    } else if(strcmp(str, "RE") == 0) { // apakah str itu 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;

}