A masala uchun yechim:

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve() {
    int n, ans = 0, a, b;
    cin >> n;
    while (n --) {
        cin >> a >> b;
        ans += (a<24 && b>10000);
    }
    cout << ans;
}
signed main() {
    ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    int t = 1;
    // cin >> t;
    while (t --) {
        solve();
        cout << '\n';
    }
    return 0;
}

B masala uchun yechim:

shunchaki har bir tangadan nechtadan berishimiz kerakligini hisoblasak yetarli edi.

n = int(input())
ans = n // 10
n -= ans * 10
ans += n // 5
n -= (n // 5) * 5
ans += n // 3
n -= (n // 3) * 3
ans += n
print(ans)

C masala uchun yechim:

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve() {
    int n, q = 0;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; ++ i) {
        cin >> arr[i];
        if (arr[i]%2) q ++;
    }
    if (q == 0) cout << -1;
    else {
        int son;
        sort(arr.begin(), arr.end());
        for (int i = n - 1; i >= 0; -- i) {
            if (arr[i]%2) {
                if (q > 1) cout << arr[i];
                else {
                    son = arr[i];
                }
                q --;
            }
            else cout << arr[i];
        }
        cout << son;
    }
}
signed main() {
    ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    int t = 1;
    // cin >> t;
    while (t --) {
        solve();
        cout << '\n';
    }
    return 0;
}

D masala uchun yechim.
Bu yerda shunchaki shaharlar orasidagi maksimal masofa bomba yo`q qila oladigan rasiusdan kichik yoki yo`qligini tekshirishimiz kerak.
 

def solve(n, m, q, arr):
   arr.append(0)
   arr.append(n)
   arr.sort()
   jav = 0
   for i in range(1, len(arr)):
       jav = max(jav, arr[i] - arr[i - 1])
   if jav <= m:
       return "No"
   else:
       return "Yes"
n, m = map(int, input().split())
q = int(input())
arr = list(map(int, input().split()))
print(solve(n, m, q, arr))

E masala uchun yechim:

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve() {
    int n;
    cin >> n;
    vector<int> arr(n);
    int ans = 0, son = 0;
    for (int i = 0, x; i < n; ++ i) {
        cin >> x;
        son = max(son + x, x);
        ans = max(ans, son);
    }
    cout << ans;
}
signed main() {
    ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    int t = 1;
    // cin >> t;
    while (t --) {
        solve();
        cout << '\n';
    }
    return 0;
}

F masala uchun yechim: 
bu yerda shunchaki 1- va oxirgi hadlarini qo`shib 2 ga bo`lsak bo`ldi edi.

def solve(a, d, n):
   an = a + (n - 1) * d
   return (a + an) // 2
for _ in range(int(input())):
 arr = list(map(str, input().split(",")))
 a = int(arr[0])
 b = int(arr[1])
 n = int(input())
 print(solve(a, (b - a), n))

G masala uchun yechim:

bunda men ikkilik sanoq sistemasidan foydalanib barcha kombinatsiyalarni ko`rib chiqib natijani topdim ko`pchilikka qulay bo`lishi uchun python dasturlash tilidan foydalandim

n = int(input())
arr = list(map(int, input().split()))
if n == 1:
 print(arr[0])
 exit()
son = 1
sonn = bin(son)[2:]
ans = []
jav = 108941294087189
son = 1
sonn = ('0' * (n - 1)) + '1'
while len(sonn) <= n:
   son += 1
   sonn = bin(son)[2:]
   if len(sonn) < n:
       sonn = ('0' * (n - len(sonn))) + sonn
   qq = 0
   for i in sonn:
       qq += int(i)
   if qq == n:
       break
   son1 = 0
   son2 = 0
   for j in range(n):
       if sonn[j] == '0':
           son1 += arr[j]
       else:
           son2 += arr[j]
   q = abs(son1 - son2)
   if jav > q:
       jav = q
print(jav)

H masala uchun yechim:

n = int(input())
arr = list(map(int, input().split()))
bir = 0
ikki = 0
uch = 0
ans = 0
for i in arr:
   if i == 1:
       bir += 1
   if i == 2:
       ikki += 1
   if i == 3:
       uch += 1
   if i == 4:
       ans += 1
if uch >= bir:
   ans += uch
   bir = 0
else:
   ans += uch
   bir -= uch
ans += (ikki // 2)
ikki -= ((ikki // 2) * 2)
if bir + ikki > 0:
   ans += (bir+(ikki*2)+3) // 4
print(ans)

Nimadir o`rgangan bo`lsangiz hursandman!