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

#include <cstring>



using namespace std ;

int main ()

{

  int N ,x=0,y=0,z=0,c=0, i;

  cin >> N ;

string a[N];

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

cin >> a[i] ;



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

x=x+1;

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

y=y+1;

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

z=z+1;

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

c=c+1;

}

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

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

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

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

    return 0 ;

}






// Faster C++ solution for the same problem (measured speedup: 5.260320066329376x):
#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);

}
// Slower C++ solution:
#include <iostream>

#include <cstring>



using namespace std;



int main()

{

    int N; cin >> N;

    int ctr = 0;

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

    char input[6];

    do

    {

        cin >> input;

        ctr++;

        if (strcmp(input,"AC") == 0)

        {

            c0++;

        }

        else if (strcmp(input,"WA") == 0)

        {

            c1++;

        }

        else if (strcmp(input,"TLE") == 0)

        {

            c2++;

        }

        else if (strcmp(input,"RE") == 0)

        {

            c3++;

        }

    } while ( ctr < N);

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

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.21124784073984x):
#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);

}
// Slower C++ solution:
#include <bits/stdc++.h>

#include <iostream>



 using namespace std;

main () 

{

  

int n;

cin >> n;

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

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

{

    string s;

    cin >>s;

    if (s=="AC")

    c1++;

    else if (s=="WA")

    c2++;

    else if (s=="RE")

    c3++;

    else if (s=="TLE")

    c4++;

}

  cout << "AC x " << c1 << "\n"<<"WA x " << c2 << "\n"<< "TLE x " << c4 << "\n"<<"RE x " << c3 << "\n";



return 0;

    

}

// Faster C++ solution for the same problem (measured speedup: 5.25441998317579x):
#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);

}
// Slower C++ solution:
#include <iostream>

#include <bits/stdc++.h>

using namespace std;



// Common file

#include <ext/pb_ds/assoc_container.hpp>

// Including tree_order_statistics_node_update

#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>







#define pb push_back

#define F first

#define S second

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



using ll = long long;

using ld = long double;

using ul = unsigned long long;

using ii = pair<int, int>;

#define Mosta     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define ReadFile            freopen("input.in","r",stdin)

#define WriteFile           freopen("output.txt","w",stdout)

const ll mod7 = 1000000007,mod = 1000000000,OO = 0x3f3f3f3f;

const ld EPS = 1e-7;

double Pi=acos(-1);

const ld rad=Pi/180.0;

ll power(ll x, ll y, ll p)

{

    ll res = 1;

    while (y > 0)

    {

        if (y & 1)

            res = (res*x) % p;

            y = y>>1;

            x = (x*x) % p;

    }

    return res;

}

ll nCr (ll n,ll m)

{

    return(m == 0)?1:n*nCr(n-1,m-1)/m;

}



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



int main()

{

    int n;

    cin >> n;



    string s1 = "AC" , s2 = "WA" , s3 = "TLE" , s4 = "RE" , ss;

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



    while(n--) {

        cin >> ss;



        if(ss == s1)

            c0++;

        else if(ss == s2)

            c1++;

        else if(ss == s3)

            c2++;

        else

            c3++;

    }



    cout << s1 << " x " << c0 << "\n";

    cout << s2 << " x " << c1 << "\n";

    cout << s3 << " x " << c2 << "\n";

    cout << s4 << " x " << c3 << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.255528697661976x):
#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);

}
// Slower C++ solution:
#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;



int main(){ int n, ac = 0, wa = 0, tle = 0, re = 0;

    cin >> n;

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

        cin >> code;

        if(code == "AC")

            ac++;

        else if(code == "WA")

            wa++;

        else if(code == "TLE")

            tle++;

        else if(code == "RE")

            re++;

    }

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

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

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

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.257565694477567x):
#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);

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int main() {

	map <string, int> m;

	long long n;

	string s;

	cin >> n;

	while(n--){

		cin >> s;

		m[s]++;

	}

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

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

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

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

}

// Faster C++ solution for the same problem (measured speedup: 5.273525958769559x):
#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);

}
// Slower C++ solution:
#include <bits/stdc++.h>

#include<cmath>

using namespace std;



int main(){

int n;

string s1="AC",s2="WA",s3="TLE",s4="RE",m;

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

cin>>n;

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

    cin>>m;

    if(m==s1)c1++;

    else if(m==s2)c2++;

    else if(m==s3)c3++;

    else if(m==s4)c4++;



}

cout<<s1<<" x "<<c1<<endl<<s2<<" x "<<c2<<endl<<s3<<" x "<<c3<<endl<<s4<<" x "<<c4;



    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.2529039660729495x):
#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);

}
// Slower C++ solution:
#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;

}




// Faster C++ solution for the same problem (measured speedup: 5.273740160220226x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <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;

}

// Faster C++ solution for the same problem (measured speedup: 5.274133558095297x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



#define 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;

}

// Faster C++ solution for the same problem (measured speedup: 5.272097941572199x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <algorithm>

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;

}


// Faster C++ solution for the same problem (measured speedup: 5.210355256646201x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include<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;

	

}

// Faster C++ solution for the same problem (measured speedup: 5.217360541903156x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

#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;



}

// Faster C++ solution for the same problem (measured speedup: 5.248448639829263x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <bits/stdc++.h>

#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;

}


// Faster C++ solution for the same problem (measured speedup: 5.254030636705278x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define 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;

}


// Faster C++ solution for the same problem (measured speedup: 5.203922669770032x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

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

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

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

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

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <string>

using namespace std;



int main() {

	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;

}

// Faster C++ solution for the same problem (measured speedup: 5.25322064304604x):
#include<stdio.h>

#include<string.h>

int main (){

	int n;

	int AC=0;

	int WA =0;

	int TLE=0;

	int RE=0;

	char kata[100];

	

	scanf("%d", &n);

while(n--){



        scanf("%s", &kata);

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

            AC++;

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

            WA++;

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

            TLE++;

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

            RE++;

        }

    }	

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

//		scanf("%s", &kata[i]);

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

//		

//		if(kata[i]=='A'&&kata[i+1]=='C'){

//			AC++;

//			continue;

//		}else if(kata[i]=='W'&&kata[i+1]=='A'){

//			WA++;

//			continue;

//		}else if(kata[i]=='T'&&kata[i+1]=='L'&&kata[i+2]=='E'){

//			TLE++;

//			continue;

//		}else if(kata[i]=='R'&&kata[i+1]=='E'){

//			RE++;

//			continue;

//		}	

// }

//	}

	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;

}