stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;cin>>n;

    string s;

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

    int intarr[4]={};

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

    {

        cin>>s;

        if(s=="AC")

            intarr[0]++;

        else if(s=="WA")

            intarr[1]++;

        else if(s=="TLE")

            intarr[2]++;

        else intarr[3]++;

    }

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

    {

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

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
// 2nd.cpp : This file contains the 'main' function. Program execution begins and ends there.

//



#include <iostream>

#include<string>

#include<vector>

using namespace std;



int main()

{

	int n;

	cin >> n;

	int a = 0;

	int b = 0;

	int c = 0;

	int d = 0;



	vector<string> ans(n);

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

	{

		cin >> ans[i];

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

			a++;

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

			b++;

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

			c++;

		else

			d++;





	}

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

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

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

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





	return 0;



}



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

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



// Tips for Getting Started: 

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

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

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

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

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

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


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;

int main ()

{

    /*/

    int i = 0, j = 1;

    cout << i;

    int *ptr = &i;

    cout << i;



    i = 2;

    cout << i;

    *ptr = 3;

    ptr = &j;

    j = i;

    cout << i;

    *ptr = 4;

      /*/



    /*/

    int n , ctn=0 ,arr;

    cin >> n ;

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

        cin >> arr;



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



    }

    cout<<ctn <<endl;



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

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

    /*/



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

    string inp;

    cin >> n ;

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

    {

        cin >> inp;

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

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

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

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



    }

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

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

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

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





    

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



map<string, int> mp;

int main() {

    int n;

    cin >> n;

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

        string s;

        cin >> s;

        mp[s]++;

    }

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

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

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

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



    return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

char st[10];

int n;

map <string,int> mp;

int main(){

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

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

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

		cin>>st;

//		cout<<st<<endl;

		mp[st]++;

	}

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

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

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

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

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;

const int MAXN = 1e5 + 5;



int a, b, c, d;

char str[500];



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

    int T;

    scanf("%d", &T);

    while (T--) {

        scanf("%s", str);

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

            a++;

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

            b++;

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

            c++;

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

            d++;

    }

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

    return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cmath>



using namespace std;



int main()

{

  int N ;

  cin >> N ;

  string S[N] ;

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

  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;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

int main()

{

	int n, b =0, c = 0,d = 0,e = 0;

	string a;

	cin >> n;

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

		cin >> a;

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

			b++;

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

			c++;

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

			d++;

		else {

			e++;

		}

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

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

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

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

	

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//#include<bits/stdc++.h>

#include<iostream>

#include<iomanip>

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<deque>

#include<set>

#include<map>

#include<bitset>

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

#define ll long long

#define ull unsigned long long

using namespace std;

const int PI = 3.14159265;

void fast() {

	std::ios_base::sync_with_stdio(0);

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

}

int main()

{

	//freopen("file.in","r",stdin);

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

	fast();



	int t;

	cin >> t;

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

	while (t--)

	{

		string s;

		cin >> s;

		if (s == "WA") {

			wa++;

		}

		else if (s == "TLE") {

			tle++;

		}

		else if (s == "RE") {

			re++;

		}

		else {

			ac++;

		}

	}

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

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

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

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



	return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include<math.h>

#include<algorithm>

#include<map>

using namespace std;



int main()

{

    map<string, int>freq;

    int n;

    string val;

    cin >> n;



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

        cin >> val;

        freq[val] += 1;

    }



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

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

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

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

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

using namespace std;

int main()

{

    int n;

    cin >> n;

    string a[n];

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

    int x1=0,x2=0,x3=0,x4=0;

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

        cin >> a[i];

        if (a[i]==b[0]) {x1++;}

        else if (a[i]==b[1]) {x2++;}

        else if (a[i]==b[2]) {x3++;}

        else if (a[i]==b[3]) {x4++;}

    }

    cout<<"AC x "<<x1<<endl;

    cout<<"WA x "<<x2<<endl;

    cout<<"TLE x "<<x3<<endl;

    cout<<"RE x "<<x4<<endl;





    return 0;

}




// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;

int main()

{

    long long n;

    cin >> n;

    string s;

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

    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 << "\nWA x " << wa

    << "\nTLE x " << tle << "\nRE x "<< re;



return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;



int main()

{

	int n;

	string val;

	cin >> n;

	int count[4] = { 0 };

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

	{

		cin >> val;

		switch (val[0])

		{

		case 'A':

			count[0]++;

			break;

		case 'W':

			count[1]++;

			break;

		case 'T':

			count[2]++;

			break;

		case 'R':

			count[3]++;

			break;

		default:

			break;

		}

	}

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



	ios::sync_with_stdio(false);

	cin.tie();

	cout.tie();

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

	

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;



}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

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

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

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

            AC++;

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

            WA++;

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

            TLE++;

        } else if(str[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;

}

// Fastest known solution for this problem: