1582
can you give me hint B
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have you try G? g is easier than f
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M direct E solve krne gya bhai alt accnt h
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Na maine contest ni diya main s, to alt s E kra
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Please send g ? Anyone
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ans is range gcd from l+1 to r of the array b
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C , D SOLUTIONS WERE SENT✅✅✅
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#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--){
string n;
cin >> n;
long long initial_sum = 0;
int c2 = 0, c3 = 0;
for(char ch : n){
int digit = ch - '0';
initial_sum += digit;
if(digit == 2) c2++;
if(digit == 3) c3++;
}
long long target = (9 - (initial_sum %9)) %9;
bool possible = false;
// Iterate through possible b (number of 3s to replace)
for(int b=0; b<=c3; b++){
// Calculate the residual needed after replacing b 3s
long long temp = (target - 6LL * b) % 9;
if(temp < 0) temp += 9;
// Solve 2*a ≡ temp mod9
// The inverse of 2 mod9 is 5 since 2*5=10≡1 mod9
long long a = (temp * 5) % 9;
// Now, a can be a + 9k where a + 9k <= c2
// Since c2 <=1e5 and a <9, check if a <=c2
if(a <= c2){
// Additionally, check if a + 9k <=c2 for some k
// Since a <=c2, it's possible to have a <=c2
possible = true;
break;
}
}
if(possible){
cout << "YES\n";
}
else{
cout << "NO\n";
}
}
}
Codeforces C sol
CF 🔥‼️
C DONE ✅ ✅
C DONE ✅✅
D. DONE ✅✅
D. DONE ✅✅
Dmm @codercpp001
bhai ye wrong de rha h
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All code are correct bro
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CF 🔥🔥
C DONE ✅✅
C. done ✅ ✅
D1 DONE ✅ ✅
D1. DONE ✅ ✅
D1. DONE ✅✅
D1. DONE ✅✅
DMM ⏩
Any code upto d1 in 15
Any code upto d1 in 15
@codercpp001
#include<bits/stdc++.h>
#include<fstream>
using namespace std;
const int mxn = 1e5 + 5, inf = 1e9 + 7;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int n;
int a[mxn + 1];
void solve(){
cin >> n;
for(int i = 1; i <= n; i++)cin >> a[i];
int ans = inf;
for(int i = 1; i <= n; i++){
int cnt = 0;
for(int j = i + 1; j <= n; j++){
cnt += (a[j] > a[i]);
}
ans = min(ans, cnt + (i - 1));
}
cout << ans << "\n";
}
signed main(){
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int tt; cin >> tt;
while(tt--){
solve();
}
return(0);
}
B sol
If not solution please give hint. I am stuck on corner cases.
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i solved b and need c if any one will trade dm me
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F ke lie bta de bhai
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E:
#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--){
string a, b, c;
cin >> a >> b >> c;
int n = a.size();
int m = b.size();
// Initialize DP table with (n+1) x (m+1) dimensions
// Using a vector of vectors for flexibility
// Initialize all values to a large number
const int INF = 1e9;
vector<vector<int>> dp(n + 1, vector<int>(m + 1, INF));
// Base case: both a and b are empty
dp[0][0] = 0;
// Fill the DP table
for(int i = 0; i <= n; ++i){
for(int j = 0; j <= m; ++j){
if(i == 0 && j == 0) continue;
// Current position in c is (i + j -1)
int pos = i + j -1;
if(pos >= c.size()){
// This should not happen as |c| = |a| + |b|
continue;
}
// If we take a character from a
if(i > 0){
int cost = dp[i-1][j];
if(a[i-1] != c[pos]){
cost +=1;
}
dp[i][j] = min(dp[i][j], cost);
}
// If we take a character from b
if(j > 0){
int cost = dp[i][j-1];
if(b[j-1] != c[pos]){
cost +=1;
}
dp[i][j] = min(dp[i][j], cost);
}
}
}
// The answer is dp[n][m]
cout << dp[n][m] << "\n";
}
}
CF E SOLUTION
Bhai E dp h 🥲, ab to Lelo koi mujhse
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E h merepe kisi ko chiye ?
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caculate using seg tree or sprase table
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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int n;
cin >> n;
vector<ll> a(n);
ll even = 0, odd = 0;
for(int i = 0; i < n; i++) {
cin >> a[i];
if(i % 2) odd += a[i];
else even += a[i];
}
ll target = (even + odd) / n;
if((even + odd) % n != 0) {
cout << "NO\n";
return;
}
ll req_even = target * ((n + 1) / 2);
ll req_odd = target * (n / 2);
if(even == req_even && odd == req_odd) cout << "YES\n";
else cout << "NO\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while(t--) solve();
return 0;
}
B sol
#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while(t--){
string s;
cin >> s;
// Convert string to a mutable vector of integers
vector<int> digits(s.size());
for(int i=0; i<s.size(); i++) digits[i] = s[i] - '0';
// Iterate from left to right
for(int i=1; i<digits.size(); i++){
// While current digit can be swapped left to increase the left digit
while(i >=1 && digits[i] > digits[i-1] +1 && digits[i] >0){
// Perform the operation: swap digits[i] and digits[i-1], after decreasing digits[i] by 1
int temp = digits[i];
digits[i] = digits[i-1];
digits[i-1] = temp -1;
// Move one position to the left to check for further possible operations
if(i >1) i--;
else break;
}
}
// Convert digits back to string
string res = "";
for(int d : digits) res += to_string(d);
cout << res << "\n";
}
}
Codeforces D sol
CODEFORCES DIV 3 ️🔥️🔥
DM FAST
D. DONE ✅✅
D. DONE ✅✅
E. DONE ✅✅
E. DONE ✅✅
E. DONE ✅✅
DM FAST @codercpp001
CF 🔥‼️
A DONE ✅✅
B DONE ✅✅
B DONE ✅✅
C DONE ✅ ✅
C DONE ✅ ✅
C DONE ✅ ✅
Dmm @codercpp001
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<pair<long long, int>> p(n + 1);
for (int i = 1; i <= n; i++) {
cin >> p[i].first;
p[i].first += i - 1;
p[i].second = 0;
}
p[1].first = 0;
sort(p.begin(), p.end());
map<long long, bool> m;
m[n] = true;
for (int i = 2; i <= n; i++) {
if (m[p[i].first])
m[p[i].first + p[i].second] = true;
}
long long r = 0;
for (const auto& [x, y] : m)
if (y)
r = x;
cout << r << "\n";
}
}
bhai iska ye solution work nhi kr rha hgala output de rha h
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I need help in cred. Assignment
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
int mod = 1e9 + 7;
void solution() {
int t;
cin>>t;
while(t--){
int n;
cin>>n;
std::vector<int>arr(n) ;
for(int i=0; i<n; i++){
cin>>arr[i];
}
int bpnt;
bool fl = 0;
for(int i=1; i<n; i++){
if(arr[i] < arr[i-1]){
bpnt =i-1;
fl=1;
break;
}
}
int cnt = bpnt;
while(upper_bound(arr.begin() , arr.end() , arr[bpnt]) != arr.end()){
int k = upper_bound(arr.begin() , arr.end() , arr[bpnt]) - arr.begin();
arr[k] = -1;
cnt++;
}
cout<<cnt<<'\n';
}
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
solution();
return 0;
}
#include <bits/stdc++.h>
using namespace std;
void solve(){
int n;
cin >> n;
vector<pair<long long, int> > a(n + 1);
for (int i = 1; i <= n; i ++){
cin >> a[i].first;
a[i].first += i - 1;
a[i].second = i - 1;
}
a[1].first = 0;
sort(a.begin(), a.end());
map<long long, bool> dp;
dp[n] = true;
for (int i = 2; i <= n; i ++){
if (dp[a[i].first])
dp[a[i].first + a[i].second] = true;
}
long long ans = 0;
for (auto [x, y] : dp)
if (y)
ans = x;
cout << ans << "\n";
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
int tests;
cin >> tests;
while (tests --)
solve();
return 0;
}
CF 🔥🔥
B. DONE ✅✅
C. done ✅ ✅
D1 DONE ✅ ✅
D1. DONE ✅ ✅
D1. DONE ✅✅
D1. DONE ✅✅
DMM ⏩
@codercpp001