80A - Panoramix's Prediction

80A - Panoramix's Prediction

#include<iostream>
using namespace std;
int primechecker(int n){
 int i;
 bool f =true;
 for(int i=2;i<n ;i++)
    if(n%i==0) {  f = false;  }
   if(f) return 1;
   else return 0;

    }
int main()
  {
   int n,m,result;
   cin>>n>>m;
    for(int i=0;i<50;i++){
   n++;
   if(primechecker(n)==1) { result = n; break;}

   }
   if(result==m) cout<<"YES";
   else cout<<"NO";
   }

104A - Blackjack

104A - Blackjack

1. #include<bits/stdc++.h>
2. using namespace std;
3. int main(){
4. int n;
5. cin>>n;
6. int d=n-10;
7. if(d==10 ) cout<<15;
8. else if(d==0) cout<<0;
9. else if(d>11 or n<=10) cout<<0;
10. else if(d<=11)cout<<4;
11. }

609A - A. USB Flash Drives

609A - USB Flash Drives

1. #include<bits/stdc++.h>
2. using namespace std;
3. int main(){
4. int n,m;
5. cin>>n>>m;
6. int a[n];
7. for(int i=0;i<n;i++) cin>>a[i];
8. sort(a,a+n);
9. reverse(a,a+n);
10. int s=0;
11. int i=0,ans=0;
12. while(1){
13. s=s+a[i];
14.  
15. ans++;i++;if(s>=m) break;
16. }cout<<ans;
17. }

870A - Search for Pretty Integers

link 870A - Search for Pretty Integers




#include<bits/stdc++.h>
using namespace std;
int main(){
 int n,m;
 cin>>n>>m;
 int a[n],b[m];
   for(int i =0;i<n;i++) cin>>a[i];
   for(int i =0;i<m;i++) cin>>b[i];
   int ans=999999;
   for(int i=0;i<n;i++){
   for(int j=0;j<m;j++){
   if(a[i]== b[j]){
   ans = min(ans, a[i] );
   }
   else {
   ans = min(ans,10*a[i]+b[j]);
   ans = min(ans,10*b[j]+a[i]);
   }
 
   }
 
   } cout<<ans;
 
}

898A - Rounding

                                                                898A - Rounding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, Vasya considers it already rounded.

For example, if n = 4722 answer is 4720. If n = 5 Vasya can round it to 0 or to 10. Both ways are correct.

For given n find out to which integer will Vasya round it.

Input

The first line contains single integer n (0 ≤ n ≤ 109) — number that Vasya has.

Output

Print result of rounding n. Pay attention that in some cases answer isn't unique. In that case print any correct answer.

Examples

input

Copy

5

output

Copy

0

input

Copy

113

output

Copy

110

input

Copy

1000000000

output

Copy

1000000000

input

Copy

5432359

output

Copy

5432360

Note

In the first example n = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.

1. #include<iostream>
2. using namespace std;
3. int f(long n){
4. int i;
5. while(n>0){
6. n = n%10;
7. return n;
8.  
9.  
10. }
11. }
12. int main(){
13. long n,in,dec;
14. cin>>n;
15. in = n;
16. dec = in;
17. long x,y;
18. long a =0,b=0;
19. while(1){
20. if(f(in)==0 ) { x=in;break;}
21. in++;
22. a++;
23. }
24. while(1){
25. if(f(dec)==0 or dec==0 ) { y=dec;break;}
26. dec--;
27. b++;
28. }
29. if(a<b) cout<<n+a;
30. else cout<<n-b;
31.  
32.  
33. return 0;
34. }

805A - Fake NP

805A - Fake NP

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.

You are given l and r. For all integers from l to r, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.

Solve the problem to show that it's not a NP problem.

Input

The first line contains two integers l and r (2 ≤ l ≤ r ≤ 109).

Output

Print single integer, the integer that appears maximum number of times in the divisors.

If there are multiple answers, print any of them.

Examples

input

Copy

19 29

output

Copy

2

input

Copy

3 6

output

Copy

3

Note

Definition of a divisor: https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html

The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.

The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.

#include<bits/stdc++.h>
using namespace std;
int main(){
 long int l,r;
 cin>>l>>r;
 
 int d=abs(l-r);
 if(d>0) cout<<2;
 else if(d==0){cout<<l; }
return 0;
}
 

                                                           798A - . Mike and palindrome

             memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

 Solution:

                                                            A. Mike and palindrome

             memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Examples

input

Copy

abccaa

output

Copy

YES

input

Copy

abbcca

output

Copy

NO

input

Copy

abcda

output

Copy

YES

 #include<bits/stdc++.h>

using namespace std;

int main(){

  string s;

  cin>>s;

int ans=0;

  int d=s.length()/2  ;

  //s.reverse( );

 int i=0;

 while(d-- )

{

  if(s[i]!=s[s.length()-1-i])

  ans++;

  i++;

}if(ans==0 and s.length()%2==1) cout<<"YES";

else if(ans==1  ) cout<<"YES";

else cout<<"NO";

 //cout<<ans;

}

798A - Mike and palindrome

                                                           798A - . Mike and palindrome

             memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

 Solution:

                                                            A. Mike and palindrome

             memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Examples

input

Copy

abccaa

output

Copy

YES

input

Copy

abbcca

output

Copy

NO

input

Copy

abcda

output

Copy

YES

 #include<bits/stdc++.h>

using namespace std;

int main(){

  string s;

  cin>>s;

int ans=0;

  int d=s.length()/2  ;

  //s.reverse( );

 int i=0;

 while(d-- )

{

  if(s[i]!=s[s.length()-1-i])

  ans++;

  i++;

}if(ans==0 and s.length()%2==1) cout<<"YES";

else if(ans==1  ) cout<<"YES";

else cout<<"NO";

 //cout<<ans;

}