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

#include<cstdlib>



/*If you are going to try,keep going until the end ,or don't even try to start*/



/* Remember, Hope is a good thing... May be the best of things... And No Good thing ever Dies !!!*/



using namespace std;



#define ll long long

#define nada "\n"

#define st string

#define S  second

#define F  first

#define pb push_back

#define MP make_pair

#define isnttriangle(a,b,c)  (a+b<c||a+c<b||b+c<a)

#define todegree(rad) rad*(180.0/pi)

#define rrp(i, a, b)  for(ll i = a; i >= b; i--)

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

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

#define rep1(i,n)     for (int i = 1; i <=n; i++)

#define allr(v)       ((v).rbegin()), ((v).rend())

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

#define fx(n)         cout<<fixed<<setprecision(n);

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

#define Test          ll t;cin>>t;while(t--)

#define mod           1000000009LL





const double ME = 2.71828;

const int    N  = 3e5 + 20;

const int   inf = 0x3f3f3f3f;

const double pi = 3.141592653;









void LOSER()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



void file()

{

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

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

}



ll gcd(ll a, ll b)

{

    //Eculidean Algorthim

    return (!b) ? a : gcd(b, a % b);

    //complixty=log(ab);

}



ll lcm(ll a,ll b)

{

    return a / gcd(a, b) * b;

}



ll Fib_recur(ll n)

{

    vector<ll>v(n+5,0);

    //BaseCase

    v[1]=0;

    v[2]=1;

    repp(i,3,n)v[i]=v[i-1]+v[i-2];

    return v[n];

}



bool isprime(int n)

{

    for(int i = 2; i <= sqrt(n); i++)

    {

        if(n%i == 0)

            return false;

    }

    return true;

}



void Solve()

{



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

  st s;

  cin>>n;

  while(n--)

  {

      cin>>s;

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

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

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

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

  }

  cout<<"AC"<<" x "<<AC<<nada;

  cout<<"WA"<<" x "<<WA<<nada;

  cout<<"TLE"<<" x "<<TLE<<nada;

  cout<<"RE"<<" x "<<RE<<nada;

}











int main()

{

    // fx(7)

    //file();

    LOSER();

    //Test



    Solve();















}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

int main()

{

	int N, counter_ac = 0, counter_wa = 0, counter_tle = 0, counter_re = 0;

	cin >> N;

	string s;

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

	{

		cin >> s;

		if (s == "AC")

		{

			counter_ac++;

		}

		else if (s == "WA")

		{

			counter_wa++;

		}

		else if (s == "TLE")

		{

			counter_tle++;

		}

		else if (s == "RE")

		{

			counter_re++;

		}

	}

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

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

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

	cout << "RE x " << counter_re << endl;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

#include <string>

#include <algorithm>

#include <cmath>

using namespace std;

int main()

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

   // your code goes here

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

    string s;

    cin>>t;

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

    {

        cin>>s;

        if (s=="AC")c0++;

        else if (s=="WA")c1++;

        else if (s=="TLE")c2++;

        else if (s=="RE")c3++;

    }

    cout<<"AC x "<<c0<<endl;

    cout<<"WA x "<<c1<<endl;

    cout<<"TLE x "<<c2<<endl;

    cout<<"RE x "<<c3<<endl;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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



using namespace std;



int main()

{

    int n,ac(0),wa(0),tle(0),re(0); cin>>n;

    string x[n];

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

        cin>>x[i];

        if (x[i]=="AC") ac++;

        else if (x[i]=="WA") wa++;

        else if (x[i]=="TLE") tle++;

        else re++;

    }

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

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;



int n,c[4];

char s[1010];



int main()

{

	scanf("%d",&n);

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

	{

		scanf("%s",s);

		if(s[0]=='A'&&s[1]=='C'&&strlen(s)==2) c[0]++;

		if(s[0]=='W'&&s[1]=='A'&&strlen(s)==2) c[1]++;

		if(s[0]=='T'&&s[1]=='L'&&s[2]=='E'&&strlen(s)==3) c[2]++;

		if(s[0]=='R'&&s[1]=='E'&&strlen(s)==2) c[3]++;

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n",c[0],c[1],c[2],c[3]);

	return 0;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

int main()

{

	int n;char str[20];

	scanf("%d", &n);

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

	while(n--)

	{

		scanf("%s", str);

		switch(str[0]){

			case'A':a++; break;

			case'W':b++;break;

			case'T':c++;break;

			case'R':d++;break;

		}

	}

	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() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

int ac,wa,re,tle;

int main(){

	int n;

	char s[8];

	scanf("%d",&n);

	while(n--){

		scanf("%s",s);

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

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

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

		if(s[0]=='R') 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() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;



int main()

{

    int N;

    cin >> N;

    string S[100000];

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

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

        cin >> S[i];

    }

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

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

            A ++;

        }

    }

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

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

            W ++;

        }

    }

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

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

            T ++;

        }

    }

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

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

            R ++;

        }

    }

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

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

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

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

    return 0;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

int n,ans[5];

string s;

int main()

{

cin>>n;

memset(ans,0,sizeof(ans));

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

	{

	cin>>s;

	if(s[0]=='A')ans[1]++;

	if(s[0]=='W')ans[2]++;

	if(s[0]=='T')ans[3]++;

	if(s[0]=='R')ans[4]++;

	}

cout<<"AC x "<<ans[1]<<endl;

cout<<"WA x "<<ans[2]<<endl;

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

cout<<"RE x "<<ans[4]<<endl;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

typedef long long ll;

map<string,int> mp;

int main(){

	std::ios::sync_with_stdio(false);

	cin.tie(0);

	int n;

	cin>>n;

	string s;

	mp["AC"] = 0; mp["WA"] = 0; mp["RE"] = 0; mp["TLE"] = 0;

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

		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;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

int n;

char s[10];

int cnta,cntw,cntt,cntr;

int main()

{

	cnta = cntw = cntt = cntr = 0;

	scanf("%d", &n);

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

	{

		scanf("%s",s);

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

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

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

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

	}

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

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

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

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

	return 0;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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



using namespace std;



int main() {

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

    int i;

    cin >> i;

    while (i--) {

        string s;

        cin >> s;

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

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

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

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

    }

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

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

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

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

    return 0;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//	Id:lys2002

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

#include<bits/stdc++.h>

using namespace std;



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

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



typedef double db;

typedef long long ll;



const ll md=1000000007;



inline void Fio(){

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

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

}

map <string,int> mp;

int main(){

    cin.tie(0);

    ios::sync_with_stdio(false);

    int n;

    cin>>n;

    string str;

    rep(i,0,n){

        cin>>str;

        if(!mp.count(str))

                mp[str]=0;

        mp[str]++;

    }

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

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

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

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

    return 0;

  }


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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


#include<iostream>

#include<string.h>

#include<algorithm>



using namespace std;







int main()

{

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

	cin>>n;

	string s;

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

	{

		cin>>s;

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

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

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

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

	}

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

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

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

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

	return 0;

}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

#include <bits/stdc++.h>



using namespace std;



int main()

{

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

    cin >> N;

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

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

    {

        cin >> x;

        if(x == AC)

            acCnt++;

        else if(x == WA)

            waCnt++;

        else if(x == TLE)

            tleCnt++;

        else if(x == RE)

            reCnt++;

    }

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

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

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

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

    return 0;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

struct test {

string T;

int n;



};

int main(){

    test test_cases[4];

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

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

    int m;

    cin>>m;

    string s;

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

        cin>>s;

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

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

                test_cases[j].n++;

            }

        }

    }

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

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

    }

    return 0;

}




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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;



int main()

{











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

cin>>N;

string s[N];



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

{ cin>>s[i];

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

    C0++;

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

        C1++;

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

        C2++;

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

        C3++;



}



cout<<"AC x "<<C0<<endl;

cout<<"WA x "<<C1<<endl;

cout<<"TLE x "<<C2<<endl;

cout<<"RE x "<<C3<<endl;

    return 0;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

#include<vector>

#include <cstdlib>

#include<cmath>

#include<string>

#include <iomanip>

#include <sstream>



using namespace std;







int main (){



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

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







    int n,i,j,cnt=0;

    cin>>n;



    string words[n];



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



        cin>>words[i];



    }



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



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



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



                cnt++;

            }



        }



        a[i]=cnt;

        cnt=0;



    }





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



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

    cout<<endl;



}













}

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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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



using namespace std;



int main()

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

cin>>x;

string arr[x];

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

    cin>>arr[i];

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

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

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

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

}

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



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



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



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





    return 0;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

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

using namespace std;

typedef long long int ll;

const int N=3e5+9;





int main()

{



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    int n;

    string s;

    map<string,int>cnt;

    cin>>n;

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

        cin>>s;

        cnt[s]++;

    }

    cout<<"AC x "<<cnt["AC"]<<'\n';

    cout<<"WA x "<<cnt["WA"]<<'\n';

    cout<<"TLE x "<<cnt["TLE"]<<'\n';

    cout<<"RE x "<<cnt["RE"]<<'\n';







    return 0;

}


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



int main() {

	long int Zx = 0, Xy = 0, Yz = 0, xyz = 0, count;

	scanf("%ld", &count);

	for (int u = 0; u < count; u++) {

		char kata[5];

		scanf("%s", &kata);

		switch (kata[0]) {

			case 'A':

				Zx++;

				break;

			case 'W':

				Xy++;

				break;

			case 'T':

				Yz++;

				break;

			case 'R':

				xyz++;

				break;		

		}	

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", Zx, Xy, Yz, xyz);

	return 0;

}

// Fastest known solution for this problem: