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Codeforces Solution

Codeforces Bear and Big Brother solution

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Codeforces Problem 791/A. Bear and Big Brother solution with CPP 100% Accepted

Before seeking assistance, it is advisable to make an effort to solve the problem yourself. This can involve researching the problem online, reviewing relevant documentation, and attempting to apply your knowledge to find a solution. By trying to solve the problem first, you can gain a better understanding of the issue and also improve your problem-solving skills. It also demonstrates to others that you have made an effort to solve the problem before seeking help.

Codeforces Problem 791/A. Bear and Big Brother solution with CPP

Question link

Code explanation

The given program takes two integer inputs ‘a’ and ‘b’ from the user, then runs a loop that repeats until ‘a’ becomes greater than ‘b’. Inside the loop, the values of ‘a’ and ‘b’ are updated by multiplying ‘a’ by 3 and ‘b’ by 2 respectively. The loop also contains an if statement to check if ‘a’ is greater than ‘b’, if so, the loop terminates using the break statement. The loop counter ‘i’ keeps track of the number of iterations it takes for ‘a’ to become greater than ‘b’. Finally, the program outputs the value of ‘i’.

Bear and Big Brother

#include<iostream>

using namespace std;
int main()
{
int a,b,i;
cin>>a>>b;

for(i=1;;i++){
a=a*3; b=b*2;
if(a>b)

break;
}

cout<<i;

}

Codeforces 236A Boy or Girl Solution

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Codeforces 236A Boy or Girl Solution || Codeforces Solution 236A

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Codeforces 236A Boy or Girl Solution in C/CPP

Question:

A. Boy or Girl

Those days, many boys use beautiful girls’ photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.

But yesterday, he came to see “her” in the real world and found out “she” is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users’ genders by their user names.

This is his method: if the number of distinct characters in one’s user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.

Input

The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.

Output

If it is a female by our hero’s method, print “CHAT WITH HER!” (without the quotes), otherwise, print “IGNORE HIM!” (without the quotes).

Note

For the first example. There are 6 distinct characters in “wjmzbmr”. These characters are: “w”, “j”, “m”, “z”, “b”, “r”. So wjmzbmr is a female and you should print “CHAT WITH HER!”.

Codeforces 236A. Boy or Girl Solution in C

#include<stdio.h>
#include<string.h>
#include<ctype.h>
int compare(const void *a, const void *b) {
return (*(char *)a - *(char *)b);
}int main() {
char a[100];
int count = 0;
scanf("%s", a);
int len = strlen(a);
qsort(a, len, sizeof(char), compare);
for (int i = 0; i < len; i++) {
if (a[i] != a[i + 1]) {
count++;
}
}
if (count % 2 == 0) {
printf("CHAT WITH HER!\n");
} else {
printf("IGNORE HIM!\n");
}
return 0;
}

Codeforces 236A in C++

#include<iostream>

#include<algorithm>

#include<string>

using namespace std;
int main()
{
string a;
int count=0;
cin>>a;
sort(a.begin(),a.end());
for(int i=0;i<a.size();i++)
{
if(a[i]!=a[i+1])
{
count++;
}
}
if(count%2==0){
cout<<"CHAT WITH HER!"<<endl;
}
else{
cout<<"IGNORE HIM!"<<endl;
}
}

Codeforces 118A. String Task solution in C/CPP

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Codeforces 118A. String Task solution in C/CPP

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118A. String Task problem link

Codeforces 118A. String Task solution in C

#include<stdio.h>
int main()
{
int i;
char s[101];
gets(s);
strlwr(s);
for(i=0;s[i]!='\0';i++)
{
if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u'||s[i]=='A'||s[i]=='E'||s[i]=='O'||s[i]=='I'||s[i]=='U'||s[i]=='y'||s[i]=='Y')
{
continue;
  }
    else {
          printf(".%c",s[i]);
    }

    }
    printf("\n");
    return 0;
}

Codeforces 118A. String Task solution in CPP

#include <iostream>

 #include <cstring>

 #include <cctype>

using namespace std;

int main() { 

string s; 

getline(cin, s);

 for (char &c : s) c = tolower(c);

 for (char c : s) { if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'y')

 { 

continue;

 } 

else { cout << "." << c; } 

}

 cout << endl; 

return 0; 

}

Codeforces 236A. Boy or Girl  Solution

Codeforces Solution

Word capitalization codeforces solution

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Word capitalization codeforces solution

Word capitalization codeforces solution

Codeforces 281A. Word Capitalization Solution in C

#include <stdio.h>
#include <string.h>
int main() {

char s[101];
 scanf("%s", s);

  s[0] = toupper(s[0]);
 printf("%s\n", s);

 return 0;
 }

Codeforces 281A. Word Capitalization Solution in CPP

#include<iostream>
#include<string>
using namespace std;
int main()
{
string s;
cin>>s;
s[0]=towupper(s[0]);
cout<<s;
}

Codeforces 791/A. Bear and Big Brother

Codeforces Solution

Petya and Strings Codeforces Solution 2024

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petya and strings codeforces solution

Question

Petya and Strings Codeforces Solution

petya and strings codeforces solution in C

#include<stdio.h>

int main()
{
int a,i;
char s1[101];
char s2[101];
gets(s1);
gets(s2);

strlwr(s1);
strlwr(s2);
int l=strcmp(s1,s2);
printf("%d\n",l);
return 0;

}
petya and strings codeforces solution
Codeforces

Codeforces 112A. Petya and Strings Solution in CPP

#include <iostream>

 #include <cstring>

 #include <cctype>

using namespace std;

int main() { 

string s1, s2;

 getline(cin, s1); 

getline(cin, s2);

for (int i = 0; i < s1.length(); i++)

 { s1[i] = tolower(s1[i]); } 

for (int i = 0; i < s2.length(); i++) 

{ s2[i] = tolower(s2[i]); }

int result = s1.compare(s2);

 cout << result << endl; 

return 0;

 }

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Beautiful Matrix Codeforces Solution

Codeforces Solution

Beautiful Matrix Codeforces Solution

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Beautiful Matrix Codeforces Solution

Question

Beautiful Matrix Codeforces Solution

Problem Specification


The goal of Codeforces Problem 263A is to determine the fewest moves required to convert a given 5×5 matrix into a “beautiful” matrix. A “beautiful” matrix is one in which the number 1 is in the center and the remaining elements are placed in a 3×3 grid around the center.

Understanding the Issue || Beautiful Matrix Codeforces Solution

Let’s start with a thorough understanding of the problem needs before diving into the solution. We are given a 5×5 matrix and must determine the smallest number of moves required to place the number 1 in the middle of the matrix. A move is defined as replacing the number 1 with an adjacent (horizontal or vertical) element. The goal is to create a “beautiful” matrix by arranging the other elements in a 3×3 grid around the number 1.

Solution in C

#include<stdio.h>

int main()
{
int i,j,x=0;
for(i=1;i<=5;i++){
for(j=1;j<=5;j++){
scanf("%d",&x);
if(x==1){
printf("%d\n",abs(i-3)+abs(j-3));
}

}

}
return 0;
}

Codeforces 282A. bit++ solution in C


Codeforces Solution

Bit++ codeforces solution in c

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Bit++ Codeforces solution in c

Question

Introduction


C programming is the foundation of modern software development. Whether you want to be a programmer or just want to expand your problem-solving skills, learning C will definitely empower you. Our step-by-step guide will take the mystery out of the process of creating a calculator, making it a fun and rewarding experience for you.

#include<stdio.h>

int main()
{
int a,i;
int x=0;
char s[4];
scanf("%d",&a);
for(i=0;i<a;i++){
scanf("%s",&s);//bit++ codeforces solution
if(s[1]=='+'){
x++;
}
else {
x--;
}
}
printf("%d\n",x);
return 0;
}

Understanding the law


Before diving into the code, let’s understand its logic. The code snippet provided is a simple calculation of a series of input characters and performs addition (+) and subtraction (-) operations accordingly. The final result is then printed on the board.[bit++ codeforces solution in c]

How does a calculator work?


The calculator uses a loop to read the specified amount of input. Each input can be an addition or subtraction function. The variable x is increased for each addition, and x is decreased for each subtraction. The final value of x is the result of the numerator.

Let’s break the rules

include: This line includes the standard input/output library, which enables input and output operations in C .

int main(): The main function is the entry point for the function.

Two integer variables, int a, i;: a and i, are declared. which will contain the number of operations the user wants to perform.

int x = 0;: the variable x is initialized to zero, which will store the last result.

char s[4];: An s character set is designed to hold the input functions. The array size is set to 4 to accommodate operators and null terminations.

scanf(“%d”, &a);: Users are requested to enter the number of operations they want to run.

for(i = 0; i < a; i++): A for loop is used to iterate over a specified number of operations.

scanf(“%s”, &s);: The user is requested to enter a function as a string.

if(s[1] == ‘+’): The code checks that the second character of the s string is ‘+’, indicating one more event.

x++;: If the condition is true, x is incremented.

else: If the condition is false (i.e. subtraction operation).

x–;: x decreases.

printf(“%d\n”, x);: The last x value (result) is printed on the screen.//bit++ codeforces solution in c

return 0;: Special interest

Codeforces 263A. Beautiful Matrix solution in C/CPP

Codeforces Solution

Domino Piling solution in C | Codeforces 50A. Domino piling solution

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Domino Piling Solution in C | Codeforces 50A. Domino piling solution

Question

A. Domino piling

You are given a rectangular board of M × N squares. Also you are given an unlimited number of standard domino pieces of 2 × 1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:

1. Each domino completely covers two squares.

2. No two dominoes overlap.

3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.

Find the maximum number of dominoes, which can be placed under these restrictions.

Input:

In a single line you are given two integers M and N — board sizes in squares (1 ≤ M ≤ N ≤ 16).

Output:

Output one number — the maximal number of dominoes, which can be placed.

Examples:

input:

2 4

output:

4

input:

3 3

output:

4

Domino piling solution in C

#include<stdio.h>

int main()
{
int a,b,p=0;
scanf("%d%d",&a,&b);
p=(a*b)/2;
printf("%d\n",p);
return 0;
}

Domino piling solution in C++

#include<iostream>
using namespace std;
int main()
{
int a,b,p=0;
cin>>a>>b;
p=(a*b)/2;
cout<<p<<endl;
return 0;
}

Next problem: Codeforces 263A. Beautiful Matrix solution in C/CPP

Codeforces Solution

Codeforces next round solution in C,Python || Codeforces 158A. Next Round Solution

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Codeforces next round solution in C,Python

QUESTION

158A. Next Round Solution solution in C

#include<stdio.h>
int main()
{
int n,i,k,count=0;
int a[100];
scanf(“%d%d”,&n,&k);

for(i=0;i<n;i++){
scanf(“%d”,&a[i]);}
k=k-1;
for(i=0;i<n;i++){

if(a[i]>=a[k]&&a[i]>0){
    count++;
}
}

 printf("%d\n",count);
 return 0;
     }

158A. Next Round Solution in Python

n,k = map(int,raw_input().split())
L = map(int,raw_input().split())
temp = L[k-1]
count=0
for i in range(0,len(L)):
    if L[i]>=temp and L[i]>0 :
        count+=1
print count

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Codeforces Solution

1A.Theatre square Codeforces Solution in C, CPP, Python || A theatre square solution

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Theatre square Codeforces Solution in C, CPP, Python

Question

Codeforces 1A.Theatre square Codeforces Solution in C

#include<stdio.h>

int main(){

long long int a,n,m,total;
scanf("%lld%lld%lld",&n,&m,&a);

double hight ,width;
hight=ceil((double)n/(double)a);
width=ceil((double)m/(double)a);//a. theatre square solution
total=hight*width;
printf("%lld\n",total);

return 0;

}

Codeforces 1A.Theatre square Solution in C++

#include <iostream>
#include <cmath>

int main() {
    long long int a, n, m, total;
    std::cin >> n >> m >> a;

    double hight, width;
    hight = ceil(static_cast<double>(n) / static_cast<double>(a));
    width = ceil(static_cast<double>(m) / static_cast<double>(a));
    total = hight * width;
    std::cout << total << std::endl;

    return 0;
}

Codeforces1A.Theatre squareSolution in Python

import math

n, m, a = map(int, input().split())

hight = math.ceil(n / a)
width = math.ceil(m / a)
total = hight * width
print(total)

Next Problem: Codeforces 1A. Theatre Square Solution in C/CPP