EKUB va EKUK :: Yechimlar

Masala A

C da itboom.h kutubxonasidagi gcd funksiyadan ham foydalansa bo’ladi.

Masala sharti

Kiritilgan ikkita sonning EKUB qiymatini hisoblaydigan dastur tuzing.

Kiruvchi ma'lumot

Har bir qatorda \(1\) dan \(2^{63}-1\) gacha bo‘lgan son

Chiquvchi ma'lumot

Ikkita sonning EKUB qiymati

#include <stdio.h>
#include <itboom.h>

long my_gcd(long n1, long n2) {
    n1 = ( n1 > 0) ? n1 : -n1;
    n2 = ( n2 > 0) ? n2 : -n2;

    long temp;
    while (n2 != 0) {
        temp = n2;
        n2 = n1 % n2;
        n1 = temp;
    }

    return n1;
}

int main() {
    long a, b;
    scanf("%lld %lld", &a, &b);
    printf("%lld", my_gcd(a, b));
    
    // yoki itboom.h dagi funksiyadan foydalanib
    // printf("%lld", gcd(a, b));

    return 0;
}

import math
import itboom

def main():
    a, b = int(input()), int(input())
    print(math.gcd(a, b))

if __name__ == '__main__':
    main()

Masala B

Masala sharti

Kiritilgan ikkita sonning EKUK qiymatini hisoblaydigan dastur tuzing.

Kiruvchi ma'lumot

Har bir qatorda \(1\) dan \(2^{32}-1\) gacha bo‘lgan son

Chiquvchi ma'lumot

Ikkita sonning EKUK qiymati

#include <stdio.h>
#include <itboom.h>

int main() {
    long a, b;
    scanf("%lld %lld", &a, &b);
    printf("%lld", a * b / gcd(a, b));
    
    //yoki itboom.h kutubxonasidan foydalanib
    //printf("%lld", lcm(a, b));

    return 0;
}

import math
import itboom

def main():
    a, b = int(input()), int(input())
    print(math.lcm(a, b))

if __name__ == '__main__':
    main()

Masala C

Masala sharti

Kiritilgan sonning tub ko‘paytuvchilari yig‘indisini hisoblaydigan dastur tuzing.

Kiruvchi ma'lumot

\(2\) dan \(2^{31}-1\) gacha bo‘lgan son

Chiquvchi ma'lumot

Kiritilgan sonnin tub ko‘paytuvchilari yig‘indisi

#include <stdio.h>
#include <itboom.h>

int main() {
    int prime_numbers[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
    
    //yoki qo'lda nechtaligini yozish kerak
    int pnn = sizeof(prime_numbers) / sizeof(int);
    

    int n, s = 0;
    scanf("%d", &n);
    
    while(n > 1) {
        for(int i = 0; i < pnn; i++) {
            int p = prime_numbers[i];
            if ((n % p) == 0) {
                s += p;
                n /= p;
            }
        }
    }

    printf("%d", s);
    return 0;
}

import math
import itboom

prime_numbers = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

def main():
    n = int(input())
    s = 0
    while n > 1:
        for p in prime_numbers:
            if n % p == 0:
                s += p
                n //= p
    
    print(s)
    
if __name__ == '__main__':
    main()

Masala D

Masala sharti

Kiritilgan son qaysi sonlarning EKUKi ekanligini aniqlaydigan dastur tuzing. Agar bir nechta sonlar juftligi mavjud bo‘lsa, juftliklar ichindan ayirmasining absolut qiymati eng kichik bo‘lgani tanlab olinsin.

Kiruvchi ma'lumot

\(1\) dan \(2^{31}-1\) gacha bo‘lgan son

Chiquvchi ma'lumot

Masala shartidagi ikkita son (qatorma-qator). Faqat kichik son birinchi qatorda, katta son ikkinchi qatorda chiqishi lozim.

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <itboom.h>

int main() {
    int n;
    scanf("%d", &n);
    
    for(int i = (int)sqrt(n); i >= 1; i--) {
        if (lcm(i, n / i) == n) {
            printf("%d\n%d", i, n / i);
            break;
        }
    }
    
    return 0;
}

import math
import itboom

def main():
    n = int(input())
    for i in range(int(math.sqrt(n)), 0, -1):
        if math.lcm(i, n // i) == n:
            print(i)
            print(n // i)
            break

if __name__ == '__main__':
    main()

Masala E

Yechimda kiritilgan belgidan 32 ni ayirib keyingi belgi chiqarilgan. ASCII kodlash jadvalida ham kichik va lotin harflari kodlari aynan 32 ga farq qiladi.

Masala sharti

Kiritilgan kichik lotin harfni katta lotin harfiga o‘tkazadigan dastur tuzing.

Kiruvchi ma'lumot

a dan z gacha bo‘lgan harf

Chiquvchi ma'lumot

Kiritilgan harfning bosh harf shakli

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