stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):




#include <bits/stdc++.h>

#include<unordered_map>

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

using namespace std;

typedef long long ll;

typedef vector<int> vi;

typedef pair<int, int> pii;

typedef vector<pii> vpii;



int main() {

    //  ios_base::sync_with_stdio(0); 

      //cin.tie(0); 

      //cout.tie(0);

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

      //freopen("output.txt", "w", stdout);

    int T = 1;

    vector<string> v = { "AC", "WA", "TLE", "RE" };

    char s1[5];

    int freq[5] = {};

   



      scanf("%d", &T);

    while (T--) {



        scanf("%s", &s1);

       string s(s1);

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

       {

           if (s == v[i])

               freq[i]++;

       }

      

    }

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

    {

        printf("%s x %d\n", v[i].c_str(), freq[i]);

    }





    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>

#include <cstring> // memset()

#include <algorithm>

#include <vector>

#include <map>

#include <set>

#include<cmath>



using namespace std;



#define ll long long

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

#define rFor(i, a, b) for (int i = (int)(a); i >= (int)(b); i--)

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

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

#define fill(a, b) memset(a, b, sizeof(a))

#define IO ios_base::sync_with_stdio(0), cin.tie(0)



const int MAX = 2e2 + 5;



int main() {

    IO;

    int n, cnt[4]{};

    string str, type[4]{"AC", "WA", "TLE", "RE"};

    cin >> n;

    For(i, 0, n) {

        cin >> str;

        For(j, 0, 4) if (str == type[j]) cnt[j]++;

    }

    For(i, 0, 4) printf("%s x %d\n", type[i].c_str(), cnt[i]);

    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>



using namespace std;



int main()

{

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

    cin>>x;

    string ans;

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

    {

        cin>>ans;

        if(ans == "AC")

            AC++;

        else if(ans == "WA")

            WA++;

        else if(ans == "TLE")

            TLE++;

        else if(ans == "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 <bits/stdc++.h>



using namespace std;



int main()

{

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

    string s;

    cin>>n;

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

    {

        cin>>s;

        if(s=="AC")

        {

            a++;

        }

        else if(s=="WA")

        {

            b++;

        }

        else if(s=="TLE")

        {

            c++;

        }

        else if(s=="RE")

        {

            d++;

        }

    }

    cout<<"AC x "<<a<<endl<<"WA x "<<b<<"\nTLE x "<<c<<"\nRE x "<<d<<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;

	string w0 = "AC", w1 = "WA", w2 = "TLE", w3 = "RE";

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

	cin >> n;

	string w[n];

	

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

		cin >> w[i];

		if(w[i] == w0){

			c0++;

		}else if(w[i] == w1){

			c1++;

		}else if(w[i] == w2){

			c2++;

		}else if(w[i] == w3){

			c3++;

		}

	}

	cout << w0 << " x " << c0 << endl;

	cout << w1 << " x " << c1 << endl;

	cout << w2 << " x " << c2 << endl;

	cout << w3 << " x " << c3 << 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>

#include <string>

#include <map>



int main()

{

    int n;

    std::map<std::string,int> mp;



    std::cin>>n;

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

    {

        std::string tmp;

        std::cin>>tmp;

        mp[tmp]++;

    }

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

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

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

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

    return 0;

}

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

#include <string.h>



int n;

char s[100000][4];

int ans[4]={0};//AC WA TLE RE



int main()

{

    scanf("%d",&n);

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

        scanf("%s",s[i]);

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

    if(strcmp(s[i],"AC")==0) ans[0]++;

    else if(strcmp(s[i],"WA")==0) ans[1]++;

    else if(strcmp(s[i],"TLE")==0) ans[2]++;

    else ans[3]++;

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

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

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

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

    return 0;

}

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

#include <string>

using namespace std;

string arr[100000];

int main()

{

    int n=0,ACs=0,TLEs=0,WAs=0,REs=0;

    cin>>n;

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

        cin>>arr[i];

    }

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

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

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

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

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

    }

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

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

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

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





    return 0;

}


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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

// 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;

    string s;

    cin>>n;

    while(n--)

    {

        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() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include <cstdio>

#include <iostream>

#include <algorithm>

using namespace std;



map <string, int> m;

int main() {

	int n;

	cin >> n;

	string s;

	for (int i = 1; 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;

	return 0;

}

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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

using namespace std;



#define ll long long



int main()

{

    int n;

    cin>>n;

    map<string, int> mp;

    string s;

    while(n--)

    {

        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() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include <algorithm>

using namespace std;

const int Size = 105;

int Arr1 [Size],Arr2[Size];

int main()

{

    int N,c1=0,c2=0,c3=0,c4=0;

    cin >> N;

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

        string S;

        cin >> S;

        if (S == "AC") c1++;

        else if (S=="WA") c2++;

        else if (S=="TLE") c3++;

        else c4++;

    }

    cout << "AC x " << c1 << endl;

    cout << "WA x " << c2 << endl;

    cout << "TLE x " << c3 << endl;

    cout << "RE x " << c4 << endl;

}


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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include<iostream>

#include<set>

#include<string>

using namespace std;

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

string x;

int main()

{

	cin >> n;

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

	{

		cin >> x;

		if (x == "AC")

			a++;

		else if (x == "WA")

			b++;

		else if (x == "TLE")

			c++;

		else if (x == "RE")

			d++;



	}

	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() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

using namespace std;

#include <string>

int main() {

    int s;

    cin>>s;

    string a[s];

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

        cin>>a[i];

    }int z=0,x=0,n=0,m=0;

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

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

            z++;

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

            x++;

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

            n++;

        else 

            m++;

    }

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

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

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

    cout<<"RE x "<<n;



}

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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include <bits/stdc++.h>

#include <algorithm>

#include <vector>



using namespace std;



int main() {





	int n; cin>>n;

	string arr[n];

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

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

	{

		cin>>arr[i];

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

			AC++;

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

			WA++;

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

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#define pi 3.14159

using namespace std;

typedef long long LL;

const LL MOD = 1e9 + 7;

const int N = 1e5 + 7, M = 1e7, OO = 0x3f3f3f3f;

#define AC ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); cin.sync_with_stdio(0);

int main()

{

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

scanf("%d",&n);

while(n--){

    string s;

    cin>>s;

    if(s=="AC"){

        ac++;

    }

    else if(s=="WA"){

        wa++;

    }

    else if(s=="TLE"){

        tle++;

    }

    else if(s=="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;

}


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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include <string>

using namespace std;



int main() {

	int ac = 0;

	int wa = 0;

	int tle = 0;

	int re = 0;

	int num = 0;

	string value;

	cin >> num;

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

		cin >> value;

		if (value == "AC") {

			ac++;

		}

		else if (value == "TLE") {

			tle++;

		}

		else if (value == "WA") {

			wa++;

		}

		else if (value == "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 t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

// 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 c[n];

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

        cin >> c[i];

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

            ac++;

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

            wa++;

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

            tle++;

        else

            re++;

    }

    cout<<"AC x "<<ac<<"\n"<<"WA x "<<wa<<"\n"<<"TLE x "<<tle<<"\n"<<"RE x "<<re<<"\n";

    return 0;

}


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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

#include<string>

#include<vector>

#include<algorithm>

using namespace std;

int main()

{

    vector <string> dash={"AC", "WA", "TLE" , "RE" };

    ios_base::sync_with_stdio(0);

    int n,m;

    cin>>n;

    vector <string> v1(n);

    for(int i=0;i<n;i++)cin>>v1[i];

    for(int i=0;i<4;i++)cout<<dash[i]<<" x "<<count(v1.begin(),v1.end(),dash[i])<<endl;

    return 0;

}

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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

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

using namespace std ;

#include<string>

int main()

{

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

	cin>>n;

	string c;

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

{

	cin>>c;

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

		a++;

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

		w++;

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

		t++;

	else

		r++;

}

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

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

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

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

}


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



int main() {

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

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[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", ac, wa, tle, re);

}

// Fastest known solution for this problem: