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

using namespace std;



using ll = long long;

using ull = unsigned long long;

using ld = long double;



#define MOD 1000000007LL



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

#define rep2(i, s, e) for(ll (i) = (ll)(s);(i) < (ll)(e);(i)++)

#define repi(i, n) for(ll (i) = 0LL;(i) <= (ll)(n);(i)++)

#define repi2(i, s, e) for(ll (i) = (ll)(s);(i) <= (ll)(e);(i)++)

#define per(i, n) for(ll (i) = (ll)(n) - 1LL;(i) >= 0LL;(i)--)

#define per2(i, s, e) for(ll (i) = (ll)(s) - 1LL;(i) >= (ll)(e);(i)--)

#define peri(i, n) for(ll (i) = (ll)(n);(i) >= 0LL;(i)--)

#define peri2(i, s, e) for(ll (i) = (ll)(s);(i) >= (ll)(e);(i)--)

#define iter(i, it) for(auto &(i): (it))



template<typename T, typename U> ostream& operator<<(ostream &s, const pair<T, U> m) {

    cout << "(" << m.first << ", " << m.second << ")";

    return s;

}

template<typename T, typename U> ostream& operator<<(ostream &s, const map<T, U> m) {

    ll c = 0;

    cout << "{ ";

    iter(i, m) cout << i << (c++ == m.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const vector<T> &v) {

    cout << "{ ";

    rep(i, v.size()) cout << v[i] << (i == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const list<T> &v) {

    ll c = 0;

    cout << "{ ";

    iter(i, v) cout << i << (c++ == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}



int main(void) {

    ll N;

    cin >> N;

    map<string, ll> m;

    rep(i, N) {

        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;

    return 0;

}


// Optimized solution to the same problem (faster):
#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>



int main(){

    int i;

    int n;

    char s[100001][8];

    int ac, wa, tle, re;

    ac = wa = tle = re = 0;



    scanf("%d", &n);

    for(i = 0;i < n;i++) scanf("%s", s[i]);



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

        switch(s[i][0]){

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }



    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n",ac, wa, tle, re);



    return 0;

}


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

using namespace std;



int main(){



    int n;

    cin >> n;



    string s[100000];



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

        cin >> s[i];



    //count関数:範囲内で完全一致する個数を調べる

    //参考:https://cpprefjp.github.io/reference/algorithm/count.html

    cout << "AC x " <<count(s, s + n, "AC") << endl;

    cout << "WA x " << count(s, s + n, "WA") << endl;

    cout << "TLE x " << count(s, s + n, "TLE") << endl;

    cout << "RE x " << count(s, s + n, "RE") << endl;

}

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

using namespace std;



int main(){

    int n;

    cin >> n;



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

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

        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;

}

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

using namespace std;

#define oo (long long)1e18

#define ll long long

#define setdp memset(dp,-1,sizeof(dp))

const ll mod = 1e9+7;

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

#define PI 3.14159265359

int n,l,r,x;

    vector<int>b(200009),a(200009);

int main()

{

    map<string,int>mp;

    cin>>n;

    while(n--)

    {

        string s;

        cin>>s;

        mp[s]++;

    }

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

    {

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

    }

}


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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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



using namespace std;



int main()

{

    string s[100000];

    int i , n ;

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

    cin>>n;

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

        cin>>s[i];}

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

        if(s[i]=="AC"){ AC+=1;}

        else if (s[i]=="WA"){WA+=1;}

        else if (s[i]=="TLE"){TLE+=1;}

        else if (s[i]=="RE"){RE+=1;}

        }

    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 c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

#include<bits/stdc++.h>



using namespace std;



int main()

{

    int n;

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

    cin>>n;

    while(n--)

    {



        char s[10];

        cin>>s;

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

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

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

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



    }

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

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

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

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

    return 0;

}






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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

using namespace std;



map<string , long long>m;



int main(){

 long long n,a,b,x;

 set<string>lst;

 string s;

 cin >>n;

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

 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;

}


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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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



using namespace std;



int main()

{

    int n;

    cin>>n;

    int cnt1=0,cnt2=0,cnt3=0,cnt4=0;

    while(n--)

    {

        string s;

        cin>>s;

        if(s=="AC")

            cnt1++;

        else if(s=="WA")

            cnt2++;

        else if(s=="TLE")

            cnt3++;

        else

            cnt4++;

    }

    cout<<"AC"<<" x "<<cnt1<<endl;

    cout<<"WA"<<" x "<<cnt2<<endl;

    cout<<"TLE"<<" x "<<cnt3<<endl;

    cout<<"RE"<<" x "<<cnt4<<endl;



    return 0;

}


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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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



using namespace std;



int main()

{

 int n;

 cin>>n;

 string s;

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

 while(n--) {

    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;

}


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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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



using namespace std;



int main() {

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

   cin>>n;

   string s[n];

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

        cin >> s[i];

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

            AC++;

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

            WA++;

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

            TLE++;

        else

            RE++;

    }

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

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

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

    cout<<"RE x "<<RE;

}






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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#pragma GCC optimize ("O3")

#include<bits/stdc++.h>

using namespace std;



/// Info.

///****************************************************************************************************************************************///

/// author                : BAHAA El-DEEN OSAMA .

/// Problem name          : A - Choose Two Numbers

/// Method                : brute force

/// Online judge          : vjudge(Contest)

/// Date                  : 8/6/2020

///****************************************************************************************************************************************///



string str;

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



int main()

{



cin>>n;

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

{

    cin>>str;

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

    else if(str == "WA"){++WA;}

    else if(str == "TLE"){++TLE;}

    else if (str == "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 c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

#include<map>

#include<fstream>

#include<vector>

using namespace std;

int main()

{

	

	int t;

	string inputs;

	

	cin >> t;



	map<string, int> freq;

	freq.emplace("AC", 0);

	freq.emplace("WA", 0);

	freq.emplace("TLE", 0);

	freq.emplace("RE", 0);



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

		cin >> inputs;

		if (freq.end() != freq.find(inputs)) {

			freq.find(inputs)->second++;

		}



	}

		

	

		cout <<freq.find("AC")->first << " x " << freq.find("AC")->second<< endl;

		cout << freq.find("WA")->first << " x " << freq.find("WA")->second << endl;

		cout << freq.find("TLE")->first << " x " << freq.find("TLE")->second << endl;

		cout << freq.find("RE")->first << " x " << freq.find("RE")->second << endl;



}


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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

 

using namespace std;

 

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

#define forp(a,b,c) for(int (a)=(b); (a)<(c); (a++))

#define form(a,b,c) for(int (a)=(b); (a)>(c); (a--))

#define fori(a,b) for(auto (a):(b))

#define pb(a) push_back(a)

#define INF ((unsigned) ~0)

#define endl "\n"

typedef long long ll;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;

ll gcd(ll a, ll b) {return b ? gcd(b, a%b) : a;}

 

int main()

{

    fastIO

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

    string s;

    cin>>n;

    while(n--)

    {

        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;

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

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

    cout<<"RE x "<<re;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

#include<math.h>

#include<vector>

#include<string>

#include<algorithm>



using namespace std;



#define vi vector<int>

#define vvi vector<vi>

#define rpt(i,x) for (int i = 0; i < x; i++)

#define fti(i,f,t) for (int i = f; i <= t; i++)

#define fte(i,f,t) for (int i = f; i < t; i++)

#define rptws(i,x,s) for (int i = 0; i < x; i += s)

#define llrpt(i,x) for (long long i = 0; i < x; i++)

#define llfti(i,f,t) for (long long i = f; i <= t; i++)

#define llfte(i,f,t) for (long long i = f; i < t; i++)





int main () {

    int n;

    cin>>n;

    int arr[4] = {0};



    rpt(i, n) {

        string s;

        cin>>s;



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

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

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

        else if (s == "RE") 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;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

using namespace std;





int main()

{

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

    cin>>n;

    string s;

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

    {

        cin>>s;

        if(s=="AC")

            AC++;

        else if (s=="WA")

            WA++;

        else if (s=="TLE")

            TLE++;

        else

            RE++;

    }



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

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

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

    cout<<"RE x "<<RE;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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



using namespace std;



int n,test_count[4];

char test[5]={"AWTR"},s[4];



int main(){

    cin>>n;

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

        cin>>s;

        for (int j=0; j<5; j++)

        {

            if (s[0]==test[j])

                test_count[j]++;

        }

    }

    cout<<"AC x "<<test_count[0]<<'\n';

    cout<<"WA x "<<test_count[1]<<'\n';

    cout<<"TLE x "<<test_count[2]<<'\n';

    cout<<"RE x "<<test_count[3]<<'\n';

    return 0;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

#include <string>

using namespace std;



int counter[4];

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



int main()

{

	string s;

	

	int n = 0; 

	cin>>n;

	

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

	   cin>>s;

	   if(s.compare(answer[0]) == 0)

	       counter[0]++;

	   else if(s.compare(answer[1]) == 0)

	       counter[1]++;

	   else if(s.compare(answer[2]) == 0)

	       counter[2]++;

	   else

	        counter[3]++;

	}

	

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

		cout<<answer[i]<<" x "<<counter[i]<<endl;

	}

	return 0;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

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

using namespace std;

#define INF 0x3f3f3f3f

#define maxn 100005

#define ll long long



void DBG() {

    cerr << "]\n";

}

template <class H, class... T> void DBG(H h, T... t) {

    cerr << h;

    if (sizeof...(t)) {

        cerr << ", ";

    }

    DBG(t...);

}

#define _DEBUG

#ifdef _DEBUG

#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)

#else

#define dbg(...) 0

#endif



int tt, n, k;

map<string, int> mp;



int main() {



    cin >> tt;

    while (tt--) {

        string s;

        cin >> s;

        mp[s]++;

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", mp["AC"], mp["WA"], mp["TLE"], mp["RE"]);

    return 0;

}

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



int main() {

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

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



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

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

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

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

  return 0;

}

// Fastest known solution for this problem: