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

#include<cstdio>

using namespace std;

int main()

{

	string s;

	getline(cin,s);

//	cout<<s<<endl;

	int flag=0,len=s.length();

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

	{

		if(!flag)

		{

			if(s[i]=='C') flag=1;

		}

		if(flag==1)

		{

			if(s[i]=='F')

			{

				flag=2;

				break;

			}

		}

	}

	if(flag==2) cout<<"Yes";

	else cout<<"No";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.341803104444209x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

#include<cstdio>

#include<string>

using namespace std;

 int q,p,w[1000001],u[1000001],o;

 string a;

 int main()

 {

     cin>>a;//首先输入字符串

     for (int i=0;i<a.size();i++)//从0位开始循环

     {

         if (a[i]=='C') 

         {

             p++;

             u[p]=i+1;

         }//判断是否发现C这个东西，并统计其位置

         else if (a[i]=='F')

         {

            q++;

             w[q]=i+1;

         }

     }//判断是否发现F，同理 统计位置情况

     if (q>=1 && p>=1)

     {

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

     {

         for (int j=1;j<=q;j++)

         {

             if (u[i]<w[j])

             {

                 o=1;

             }

         }

     }//循环，找到C是否在F的前面

     if (o==1) cout<<"Yes"<<endl;//在前面则Yes

     else cout<<"No"<<endl;//否则No   

     }

     else cout<<"No"<<endl;//不存在两个数，直接No

     //以上为全部过程，希望能理解。

 }

// Faster C++ solution for the same problem (measured speedup: 5.322703874618073x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

#include<cstdio>

#include<string>

using namespace std;

string s;

int main(){

    cin>>s;

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

        if(s[i]=='C'){

            for(int j=i+1;j<s.size();j++){

                if(s[j]=='F'){

                    cout<<"Yes";

                    return 0;

                }

            }

        }

    }

    cout<<"No";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.3100802742926465x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

#include<iostream>

using namespace std;

char c[111];

int le,f;

int main()

{

    cin>>c;

    le=strlen(c);

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

    {

        if(c[i]=='C'&&f==0)

        f=1;

        if(c[i]=='F'&&f!=0)

        {

            cout<<"Yes"<<endl;

            return 0;

        }

    }

    cout<<"No"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.320748524171252x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

#include<cstdio>

#include<string>

using namespace std;

 int q,p,w[1000001],u[1000001],o;

 string a;

 int main()

 {

     cin>>a; 

     for (int i=0;i<a.size();i++) 

     {

         if (a[i]=='C') 

         {

             p++;

             u[p]=i+1;

         } 

         else if (a[i]=='F')

         {

            q++;

             w[q]=i+1;

         }

     } 

     if (q>=1 && p>=1)

     {

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

     {

         for (int j=1;j<=q;j++)

         {

             if (u[i]<w[j])

             {

                 o=1;

             }

         }

     } 

     if (o==1) cout<<"Yes"<<endl; 

     else cout<<"No"<<endl;   

     }

     else cout<<"No"<<endl; 

     

 }  

// Faster C++ solution for the same problem (measured speedup: 5.322703874618073x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

using namespace std;

int main()

{

    string s;

    cin >> s;

    int c = s.find("C"); 

    int f = s.find("F", c);

    if(c >= s.length() || f >= s.length() || c > f)

        cout << "No" << endl;

    else

        cout << "Yes" << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.366853385233142x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

}
// Slower C++ solution:
// luogu-judger-enable-o2

#include<bits/stdc++.h>

using namespace std;

string c;

int main()

{

	cin>>c;

	int n=c.size();

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

	{

		if(c[i]=='C')

		{

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

			{

				if(c[j]=='F')

				{

					cout<<"Yes";

					return 0;

				}

			}

		}

	}

	cout<<"No";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.3109035222826435x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

using namespace std;

char a[266];

int main()

{

	int b=100;

	cin>>a;

	for(int i=0;i<strlen(a)&&b!=0;i++)

	{

		if(a[i]=='C'||b==10)

		{

			b=10;

			if(a[i]=='F'&&b==10)

				b=0;

		}

	}

	if(b==0)

		cout<<"Yes";

	else

		cout<<"No";	

}

// Faster C++ solution for the same problem (measured speedup: 5.309461444863717x):
#include<cstdio>

#include<cstring>

const int N=1100;



int l;

bool f1,f2;

char ch[N];

int main(){

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

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



	ch[0]=3;

	scanf("%s",ch+1);

	l=strlen(ch)-1;

	

	f1=f2=false;

	for(int i=1; i<=l; i++){

		if(ch[i]!='C' and ch[i]!='F') continue;

		

		if(!f1 and ch[i]=='C'){

			f1=true;

			continue;

		}

		if(f1 and ch[i]=='F'){

			f2=true;

			break;

		}

	}

	

	if(f2){

		printf("Yes\n");

	}

	else{

		printf("No\n");

	}

	

	return 0;

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

using namespace std;

char ch[1314520];

int main(){

	scanf("%s",ch);

	int a=0;

	for(int i=0;i<strlen(ch);i++){

		if(ch[i]=='C'&&!a){

			a++;

			continue;

		}

		if(ch[i]=='F'&&a==1){

			printf("Yes\n");

			return 0;

		}

	}

	printf("No\n");

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.2652724797253505x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<string>

#include<cstdio>

using namespace std;

string s;

bool t;

int main(){

	cin>>s;

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

		if(t && s[i]=='F'){

			puts("Yes");

			return 0;

		}

		if(s[i]=='C')t=true;

	}

	puts("No");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.303096110405727x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include <iostream>

#include <cmath>

#include <algorithm>



using namespace std;



int main(){

    char c[101];

    cin>>c;

    //

    bool cFound = false, fFound = false;

    

    for(int i = 0;c[i] != '\0';i++){

        if(c[i] == 'C' && !fFound) cFound = true;

        if(c[i] == 'F' && cFound) fFound = true;

    }

    if(cFound && fFound) cout<<"Yes";

    else cout<<"No";

}

// Faster C++ solution for the same problem (measured speedup: 5.316308689075503x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

#include<string>

using namespace std;

string s;

int l;

int main()

{

    cin>>s;

    l=s.size();

    for(register int i=0;i<l;i++)

    {

        if(s[i]=='C')

        {

            for(register int j=i+1;j<l;j++)

            {

                if(s[j]=='F')

                {

                    cout<<"Yes";

                    return 0;

                }

            }

        }

    }

    cout<<"No";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.317893015306747x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

using namespace std;

char cf[111];



int main(){

    scanf("%s",cf);

    bool c=false,f=false;

    int l=strlen(cf);

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

    if(c==0&&cf[i]=='C')

        c=true;

    if(c==1&&cf[i]=='F')

        f=true;}

    if(c&&f)printf("Yes");

    else printf("No");

    return 0;}

// Faster C++ solution for the same problem (measured speedup: 5.267096972663625x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

using namespace std;

char a[10010]; 

bool c, f;

int main() {

    scanf("%s", a);

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;

        if(a[i]=='F'&&c==true) f=true;

    }

    if(f==true) cout<<"Yes"<<endl; 

    else cout<<"No\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.318797413519808x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

using namespace std;

int main()

{

	int i;

	string s;

	cin>>s;

	for(i=0;i<s.size();i++)

		if(s[i]=='C') break;

	for(int j=i+1;j<s.size();j++)

		if(s[j]=='F') {cout<<"Yes";return 0;}

	cout<<"No";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.314009862214981x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstring>

using namespace std;



string a,b="CF"; 

int al,bl;



int main(){

    getline(cin,a);

    al=a.length();

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

        if(a[i]==b[bl]){ 

            bl++;

            if(bl==2)   

                break;

        }

    if(bl==2) cout<<"Yes";

    else cout<<"No";

    cout<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.318679222092367x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

}
// Slower C++ solution:
// ==============================

//  author: memset0

//  website: https://memset0.cn

//  date: 2018.08.06 14:06:24

// ==============================

#include <bits/stdc++.h>

#define ll long long

using namespace std;



int len;

char s[110];



int main() {

	scanf("%s", s), len = strlen(s);

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

		if (s[i] == 'C')

			for (int j = i + 1; j < len; j++)

				if (s[j] == 'F') {

					printf("Yes\n");

					return 0;

				}

	printf("No\n");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.266913448640689x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

using namespace std;

char S[110];

int main()

{

	scanf("%s",S);

	int l=strlen(S);

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

	{

		if(S[i]=='C')

		{

			for(int j=i+1;j<l;j++)

			{

				if(S[j]=='F')

				{

					printf("Yes\n");

					return 0;

				}

			}

		}

	}

	printf("No\n");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.266920714212309x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

using namespace std;

string s;

int l;

int main()

{

    cin>>s;

    l=s.size();

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

    {

        if(s[i]=='C')

        {

            for(int j=i+1;j<l;j++)

            {

                if(s[j]=='F')

                {

                    cout<<"Yes";

                    return 0;

                }

            }

        }

    }

    cout<<"No";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.317893015306747x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

char a[10010]; 

bool c, f; 

int main() {

    scanf("%s", a); 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true; 

        if(a[i]=='F'&&c==true) f=true;

    }

    if(f==true) printf("Yes\n"); 

    else printf("No\n"); 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.265494345824261x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<iostream>

using namespace std;

int main()

{

    string s;

    int f=0;

    getline(cin,s);

    for(int i=0;i<s.length();i++)

    {

        if(s[i]=='C'&&!f)f=1;

        if(s[i]=='F'&&f)f=2;

    }

    if(f==2)cout<<"Yes"<<endl;

    else cout<<"No"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.352513507237164x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

#include<string>

using namespace std;

 int q,p,w[1000001],u[1000001],o;

 string a;

 int main()

 {

     cin>>a;//首先输入字符串

     for (int i=0;i<a.size();i++)//从0位开始循环

     {

         if (a[i]=='C') 

         {

             p++;

             u[p]=i+1;

         }//判断是否发现C这个东西，并统计其位置

         else if (a[i]=='F')

         {

            q++;

             w[q]=i+1;

         }

     }//判断是否发现F，同理 统计位置情况

     if (q>=1 && p>=1)

     {

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

     {

         for (int j=1;j<=q;j++)

         {

             if (u[i]<w[j])

             {

                 o=1;

             }

         }

     }//循环，找到C是否在F的前面

     if (o==1) cout<<"Yes"<<endl;//在前面则Yes

     else cout<<"No"<<endl;//否则No   

     }

     else cout<<"No"<<endl;//不存在两个数，直接No

     //以上为全部过程，希望能理解。

 }

// Faster C++ solution for the same problem (measured speedup: 5.330479172217456x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

using namespace std;

int main()

{

	string s;

	getline(cin,s);

//	cout<<s<<endl;

	int flag=0,len=s.length();

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

	{

		if(!flag)

		{

			if(s[i]=='C') flag=1;

		}

		if(flag==1)

		{

			if(s[i]=='F')

			{

				flag=2;

				break;

			}

		}

	}

	if(flag==2) cout<<"Yes";

	else cout<<"No";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.349963996574851x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

#include<string>

using namespace std;

 int q,p,w[1000001],u[1000001],o;

 string a;

 int main()

 {

     cin>>a;//首先输入字符串

     for (int i=0;i<a.size();i++)//从0位开始循环

     {

         if (a[i]=='C') 

         {

             p++;

             u[p]=i+1;

         }//判断是否发现C这个东西，并统计其位置

         else if (a[i]=='F')

         {

            q++;

             w[q]=i+1;

         }

     }//判断是否发现F，同理 统计位置情况

     if (q>=1 && p>=1)

     {

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

     {

         for (int j=1;j<=q;j++)

         {

             if (u[i]<w[j])

             {

                 o=1;

             }

         }

     }//循环，找到C是否在F的前面

     if (o==1) cout<<"Yes"<<endl;//在前面则Yes

     else cout<<"No"<<endl;//否则No   

     }

     else cout<<"No"<<endl;//不存在两个数，直接No

     //以上为全部过程，希望能理解。

 }

// Faster C++ solution for the same problem (measured speedup: 5.3308355880704035x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<cstdio>

#include<string>

using namespace std;

string s;

int main(){

    cin>>s;

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

        if(s[i]=='C'){

            for(int j=i+1;j<s.size();j++){

                if(s[j]=='F'){

                    cout<<"Yes";

                    return 0;

                }

            }

        }

    }

    cout<<"No";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.318192702151977x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

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

#include<iostream>

using namespace std;

char c[111];

int le,f;

int main()

{

    cin>>c;

    le=strlen(c);

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

    {

        if(c[i]=='C'&&f==0)

        f=1;

        if(c[i]=='F'&&f!=0)

        {

            cout<<"Yes"<<endl;

            return 0;

        }

    }

    cout<<"No"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.328877250354347x):
#include<cstdio>

#include<cstring>

char a[10010];//字符数组 

bool c, f;//标志变量 

int main() {

    scanf("%s", a);//读入字符串 

    for(int i=0; i<strlen(a); i++) {

        if(a[i]=='C') c=true;//找到C后标记 

        if(a[i]=='F'&&c==true) f=true;//在找到C后再找到F时标记 

    }

    if(f==true) printf("Yes\n");//如果找到C和F输出Yes 

    else printf("No\n");//否则输出No 

    return 0;

}