A. Yo'ldagi daraxtlar

\(a\) va \(b\) nuqtalar orasidagi nuqtalar soni: \(|a - b| + 1\)

B. Gugurt donalari

Tomonlari a va b bo'lgan to'g'ri to'rtburchakning diagonali uzunli \(\sqrt{a^2 + b ^ 2}\) teng. Agar gugurt donasi uzunligi \(\sqrt{a^2 + b ^ 2}\) dan kichik yok teng bo'lsa javob BOX aks holda TRASH.

C. Rasmni kattalashtirish

Pikselni enini \(b\) martaga oshiradigan bo'lsa bu piksel \(b\) marta ketma ket yoziladi. Agar qatorni bo'yini \(a\) martaga oshiradigan bo'lsak qator \(a\) marta chiqadi. Yaxshiroq tushunish uchun rasmga qarang.

D. Telefonda raqam terish

Kiritilgan telefon nomeridagi har bir raqam yonma yun tursa javob YES bo'ladi, aks holda NO.

E. Massivlar soni

Bu masala dynamic-programming bilan yordamida osonlikcha hal qilish mumkin. Keling \(dp[n][k]\) deb uzunligi \(n\) ga teng bo'lgan permutatsiyalar ichida inversiyalar soni \(k\) bo'lgan massivlar soni (boshqacharoq qilib aytganda (n, k) kiritilgandagi javob). 3 hil holat bor:
\(dp[0][k] = 0\), massiv bo'sh
\(dp[n][0] = 1\), inversiyasi soni 0 ta bo'lgan yagona massiv bor.
\(dp[n][k] = dp[n - 1][k - n + 1] + dp[n - 1][k - n + 2] + ... + dp[n - 1][k]\), shunday bo'lishini isbotlaymiz. Misol uchun \(dp[n - 1][k - n + 2]\) bo'lgan holat javobga qo'shilayabdi, lekin nega? Chunki uzunligi \((n - 1)\)bo'lgan va inversiyalar soni \((k - n + 2)\) bo'lgan permutatsiyalarni 2-o'rniga n sonini qo'shsak undagi inversiyalar soni \(k\) ta bo'lib qoladi va uzunligi ham \(n\) bo'ladi.

C++ dasturlash tilida masalalarning yechimlari:

A. Yo'ldagi daraxtlar

#include <iostream>
using namespace std;


int main() {
    int a, b;
    cin >> a >> b;
    cout << b - a + 1;

    return 0;
}

B. Gugurt donalari

#include <bits/stdc++.h>
using namespace std;


int main() {
    int n, a, b;
    cin >> n >> a >> b;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        cout << (x <= sqrt(a * a + b * b) ? "BOX\n" : "TRASH\n");
    }

    return 0;
}

C. Rasmni kattalashtirish

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < a; j++) {
            for (int k = 0; k < m; k++) {
                for (int l = 0; l < b; l++) cout << s[k];
            }
            cout << '\n';
        }
    }

    return 0;
}

D. Telefonda raqam terish

#include <iostream>
#include <vector>
using namespace std;

int main()
{
	string s;
	cin >> s;
	vector<pair<int, int>> order = {{4, 2}, {1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}};
	for (int i = 1; i < s.size(); i++)
	{
		if (abs(order[s[i - 1] - '0'].first - order[s[i] - '0'].first) +
				abs(order[s[i - 1] - '0'].second - order[s[i] - '0'].second) !=
			1)
		{
			cout << "NO";
			return 0;
		}
	}
	cout << "YES";
	return 0;
}

E. Massivlar soni

#include <iostream>
#include <vector>
using namespace std;
using ll = long long;
const int mod = 1e9 + 7;

int main() {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> dp(n + 1, vector<int>(k + 1));
    for (int i = 0; i <= n; i++) {
        ll sum = 0;
        for (int j = 0; j <= k; j++) {
            if (i == 0) continue;
            sum += dp[i - 1][j];
            if (j == 0) {
                dp[i][0] = 1;
                continue;
            }
            // dp[i][j - i] qo'shilib qolgan bo'lsa chiqarvorish kerak
            if (j - i >= 0) sum -= dp[i - 1][j - i];
            dp[i][j] = sum % mod;
        }
    }
    cout << dp[n][k];

    return 0;
}

Python dasturlash tilidagi yechimlar:

A. Yo'ldagi daraxtlar

a = int(input())
b = int(input())
print(b - a + 1)

B. Gugurt donalari

n, a, b = map(int, input().split())
for i in range(n):
    x = int(input())
    print("BOX" if x ** 2 <= (a ** 2 + b ** 2) else "TRASH")

C. Rasmni kattalashtirish

n, m, a, b = map(int, input().split())
for i in range(n):
    s = input()
    for j in range(a):
        for k in range(m):
            for l in range(b):
                print(s[k], end='')
        print()

D. Telefonda raqam terish

s = input()
order = [[4, 2], [1, 1], [1, 2], [1, 3], [2, 1],
         [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]
for i in range(1, len(s)):
    if abs(order[int(s[i - 1])][0] - order[int(s[i])][0]) + abs(order[int(s[i - 1])][1] - order[int(s[i])][1]) != 1:
        print("NO")
        exit()
print("YES")

E. Massivlar soni

TL = False
ML = False

if TL or ML:
    print("Learn cpp")
else:
    n, k = map(int, input().split())
    dp = [[0 for _ in range(k + 2)] for _ in range(n + 2)]
    mod = 10 ** 9 + 7
    for i in range(n + 1):
        sum = 0
        for j in range(k + 1):
            if i == 0:
                continue
            sum += dp[i - 1][j]
            if j == 0:
                dp[i][0] = 1
                continue
            if j - i >= 0:
                sum -= dp[i - 1][j - i]
                
            dp[i][j] = sum % mod
            
            
    print(dp[n][k])