stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

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

 cin>>n;

 string s;

 while(n--)

 {

     cin>>s;

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

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

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

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

 }

 cout<<"AC x "<<a<<"\nWA x "<<w<<"\nTLE x "<<t<<"\nRE x "<<r;



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.235395477284211x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;

int N,a,b,c,d;

string x;

int main() {

	cin>>N;

	for(int i=1;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: 5.198607913016982x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;



int main() {

     int n , arr[4] = {0} ;



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



     cin >> n ;



     while(n--){



          cin >> s ;

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



               if(s == arr_st[i])arr[i]++ ;

          }



     }

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



          cout << arr_st[j] << " x " << arr[j] << endl ;

     }

}


// Faster C++ solution for the same problem (measured speedup: 5.232806949538896x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#include <cmath>

#include <algorithm>

#include <cstdlib>

using namespace std;



int main() {

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

    scanf("%d",&x);

    string s;

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

        cin>>s;

        if(s=="AC")

            a++;

        else if (s=="WA")

        b++;

        else if(s=="TLE")

            c++;

        else if(s=="RE")

            d++;

    }

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



}

// Faster C++ solution for the same problem (measured speedup: 5.191828757005273x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;



int main()

{

    ios::sync_with_stdio(false);  cin.tie(0); 

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

    string s;

    cin >> n;

    while(n--)

    {

        cin >> s;

        if (s == "AC")

            ac++;

        else if (s == "TLE")

            tle++;

        else if (s == "WA")

            wa++;

        else

            re++;

    }

    cout << "AC x " <<ac <<"\nWA x "<<wa;

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.260637600330159x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int x;

    cin>>x;

    string arrS[x];

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

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

    {

        cin>>arrS[i];

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

            AC++;

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

            WA++;

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

            TLE++;

        else

            RE++;

    }



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

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

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

    cout<<"RE x "<<RE;









    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.241431679066915x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

}
// 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()

{

//have submited with my brother account becuase of forget to swap accounts

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: 5.191377970370541x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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



using namespace std;



int main()

{

    int n;

    cin>>n;

    string s;

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

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.241369484975064x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;

int n, arr[4];

int main()

{

	cin>>n;

	string str;

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

		cin>>str;

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

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

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

		else if(str == "RE") arr[3]++;



	}



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

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

	cout<<"TLE x "<<arr[2]<<endl;

	cout<<"RE x "<<arr[3]<<endl;



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.2420767619730855x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

}
// Slower C++ solution:


#include <bits/stdc++.h>

#include<vector>

using namespace std;

int main()

{

    int N;

    cin>>N;

    int count1=0,count2=0,count3=0,count4=0;

    string s;

    vector<string>v;

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

        cin>>s;

        v.push_back(s);

    }

    count1=count(v.begin(),v.end(),"AC");

    count2=count(v.begin(),v.end(),"WA");

    count3=count(v.begin(),v.end(),"TLE");

    count4=count(v.begin(),v.end(),"RE");

    cout<<"AC"<<" "<<"x"<<" "<<count1<<endl;

    cout<<"WA"<<" "<<"x"<<" "<<count2<<endl;

    cout<<"TLE"<<" "<<"x"<<" "<<count3<<endl;

    cout<<"RE"<<" "<<"x"<<" "<<count4<<endl;







    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.263887603222908x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;

typedef long long ll;

#define INF 0x3f3f3f3f

const int N=102,M=10010,mod=998244353;

ll n,m,T,k,cnt,x,y,z,indx,t,ans,a[N];

string s;

int main()

{

	scanf("%lld",&n);

	while (n--)

	{

		cin>>s;

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

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

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

		if (s=="RE") a[4]++;

	}

	printf("AC x %lld\n",a[1]);

	printf("WA x %lld\n",a[2]);

	printf("TLE x %lld\n",a[3]);

	printf("RE x %lld\n",a[4]);

}

// Faster C++ solution for the same problem (measured speedup: 5.195376230864469x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 5.243369338409143x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;

int i,j,num[4],n;

char s[10];

int main()

{

	scanf("%d",&n);

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

	{

		scanf("%s",s);

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

		if (strcmp(s,"WA")==0) num[1]++;

		if (strcmp(s,"TLE")==0) num[2]++;

		if (strcmp(s,"RE")==0) num[3]++; 

	}

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

}

// Faster C++ solution for the same problem (measured speedup: 5.124580069348823x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#include <cstdio>

#include <cstring>

#include <cmath>

#include <iostream>

#include <algorithm>

#include <stack>

using namespace std;



int main(void)

{

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

  int N;

  scanf("%d",&N);

  char s[100];

  int numA=0,numW=0,numT=0,numR=0;

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

  {

    scanf("%s",s);

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

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

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

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

  }

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

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

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

  printf("RE x %d",numR);





  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.122852134269812x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#define ll long long

#define endl "\n"

#define pb push_back

#define READ(FILE) freopen(FILE,"r",stdin)

#define WRITE(FILE) freopen(FILE,"w",stdout)

#define loop(i,s,e) for(int i=s;i<e;i++)

#define mego fast();

using namespace std;

void fast()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



ll lcm(ll x, ll y)

{

    return (x*y)/__gcd(x,y);

}

int sum(string s)

{

    int x=0,sz=s.size();

    loop(i,0,sz)x+=(s[i]-'0');

    return x;

}

/*

struct node

{

int data;

node *link;

};

typedef node* nodeptr;

void head_insert(nodeptr& head,int x)

{

    nodeptr tmp;

    tmp=new node;

    tmp->data=x;

    tmp->link=head;

    head=tmp;

}

nodeptr search(nodeptr head,int x)

{

    nodeptr cur=head;

    if(cur ==  nullptr)return cur;

    while(cur->data!=x&&cur->link!=nullptr)

    {

        cur=cur->link;

    }

    if(cur->data==x)return cur;

    else return nullptr;

}

void show(nodeptr head)

{

    nodeptr cur=head;

    if(cur==nullptr)return;

   // cout<<head->data<<" ";

    while(cur->link!=nullptr){

        cout<<cur->data<<" ";

        cur=cur->link;

    }

    cout<<endl;

}

*/



int main()

{

    mego

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

    string s;

    cin>>n;

    loop(i,0,n)

    {

        cin>>s;

        if(s=="AC")

            a++;

        else if(s=="WA")

            b++;

        else if(s=="TLE")

            c++;

        else

            d++;



    }

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

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

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

    cout<<"RE x "<<d;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.305595250492732x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

using namespace std;



int main()

{

    ios::sync_with_stdio(0);

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

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

    string s;

    while (n--)

    {

        cin >> s;

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

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

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

        else re++;

    }

    cout << "AC x " << ac << "\nWA x " << wa << "\nTLE x " << tle << "\nRE x " << re << '\n';

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.300156401266638x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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



using namespace std;





int main()

{

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

string stats;

cin>>n;

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

    cin>>stats;

    if(stats=="AC"){ac++;}

    else if(stats=="WA"){wa++;}

    else if(stats=="RE"){re++;}

    else if(stats=="TLE"){tle++;}



}

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: 5.242177526044379x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#include <vector>

#include <queue>

#include <algorithm>

#include <cmath>

#include <map>

#include <stdio.h>

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

#define ll long long

#define pie 3.141592653



using namespace std;





int main()

{

	IO;



	int n;

	string s;

	vector <pair<string, int>>vp;

	vp.push_back(make_pair("AC", 0));

	vp.push_back(make_pair("WA", 0));

	vp.push_back(make_pair("TLE", 0)); 

	vp.push_back(make_pair("RE", 0));

	cin >> n;



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

	{

		cin >> s;

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

		{

			if (vp[j].first == s)

			{

				vp[j].second++;

				break;

			}

		}

	}

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

	{

		cout << vp[i].first << " x " << vp[i].second;

		if (i < 3)

			cout << "\n";

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.336619977706552x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

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

#include<map>



using namespace std;



int main()

{

    int n;

    cin>>n;

    map<string,int>mymap;

    char str[100];

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

        cin>>str,mymap[str]++;

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

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

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

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



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.26496129502563x):
#include <stdio.h>

#include <string.h>

//subgoals

//1. scan semua input

//2. cari jumlah AC,WA,TLE,RE



int main (){

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

	scanf ("%d",&N);

	

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

	

	char str [255];

	scanf ("%s",str);

	

	if (strcmp(str,"AC")==0){//"AC" - "AC"=0

		AC++;

	}else if (strcmp(str,"WA")==0){ //if str is WA

		WA++;

	}else if (strcmp(str,"TLE")==0){//if str is TLE

		TLE++;

	}else if (strcmp(str,"RE")==0){//if str i 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;

}