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For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
String area = conditional(side);
System.out.println(area);
}static String conditional(int n){
if(n==1){return "one";}
else if(n==2){return "two";}
else if(n==3){return "three";}
else if(n==4){return "four";}
else if(n==5){return "five";}
else{
return "Greater than 5";}
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().toString();
int n=Integer.parseInt(s);
int ch=0;
if(n>0){
ch=1;
}
else if(n<0) ch=-1;
switch(ch){
case 1: System.out.println("Positive");break;
case 0: System.out.println("Zero");break;
case -1: System.out.println("Negative");break;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: n = input()
if '-' in list(n):
print('Negative')
elif int(n) == 0 :
print('Zero')
else:
print('Positive'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch (num > 0)
{
// Num is positive
case 1:
printf("Positive");
break;
// Num is either negative or zero
case 0:
switch (num < 0)
{
case 1:
printf("Negative");
break;
case 0:
printf("Zero");
break;
}
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(read.readLine());
long[] arr = new long[N];
StringTokenizer st = new StringTokenizer(read.readLine());
for(int i=0; i<N; i++) {
arr[i] = Long.parseLong(st.nextToken());
}
Arrays.sort(arr);
long res = 0;
for(int i=N-1;i >-1; i--) {
res += arr[i]*i;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
arr.sort(reverse=True)
l=[]
for i in range(n-1):
l.append(arr[i]*((n-1)-i))
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n+1];
for(int i=1;i<=n;++i){
cin>>a[i];
}
sort(a+1,a+n+1);
int ans=0;
for(int i=1;i<=n;++i)
ans+=(a[i]*(i-1));
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Consider the integer sequence A = {1, 2, 3, ...., N} i.e. the first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing <b>at most one</b> element from A.
A <b>subarray</b> of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains three integers: N, L, and S.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>9</sup>
1 <= L <= N-1
1 <= S <= 10<sup>18</sup> <b> (Note that S will not fit in a 32-bit integer) </b>For each testcase, print "YES" (without quotes) if a required subarray can exist, and "NO" (without quotes) otherwise.Sample Input:
3
5 3 11
5 3 5
5 3 6
Sample Output:
YES
NO
YES
Sample Explanation:
For the first test case, we can remove 3 from A to obtain A = {1, 2, 4, 5} where {2, 4, 5} is a required subarray of size 3 and sum 11., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long INF = 1e18;
const int INFINT = INT_MAX/2;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x, y) freopen(x, "r", stdin); freopen(y, "w", stdout);
#define out(x) cout << ((x) ? "YES\n" : "NO\n")
#define CASE(x, y) cout << "Case #" << x << ":" << " \n"[y]
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now(); auto end_time = start_time;
#define measure() end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
#define reset_clock() start_time = std::chrono::high_resolution_clock::now();
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll norm(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; g--; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n): adj(n+1) {}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
struct LCA
{
vll tin, tout, level;
vector<vll> up;
ll timer;
void _dfs(vector<vector<int>> &adj, ll x, ll par=-1)
{
up[x][0] = par;
REP(i, 1, 20)
{
if(up[x][i-1] == -1) break;
up[x][i] = up[up[x][i-1]][i-1];
}
tin[x] = ++timer;
for(auto &p: adj[x])
{
if(p != par)
{
level[p] = level[x]+1; _dfs(adj, p, x);
}
}
tout[x] = ++timer;
}
LCA(Graph &G, ll root)
{
int n = G.adj.size();
tin.resize(n); tout.resize(n); up.resize(n, vll(20, -1)); level.resize(n, 0);
timer = 0;
//does not handle forests, easy to modify to handle
_dfs(G.adj, root);
}
ll find_kth(ll x, ll k)
{
ll cur = x;
REP(i, 0, 20)
{
if((k>>i)&1) cur = up[cur][i];
if(cur == -1) break;
}
return cur;
}
bool is_ancestor(ll x, ll y)
{
return tin[x]<=tin[y]&&tout[x]>=tout[y];
}
ll find_lca(ll x, ll y)
{
if(is_ancestor(x, y)) return x;
if(is_ancestor(y, x)) return y;
ll best = x;
REPd(i, 19, 0)
{
if(up[x][i] == -1) continue;
else if(is_ancestor(up[x][i], y)) best = up[x][i];
else x = up[x][i];
}
return best;
}
ll dist(ll a, ll b)
{
return level[a] + level[b] - 2*level[find_lca(a, b)];
}
};
long long ap_sum(long long st, long long len) {
//diff is always 1
long long en = st + len - 1;
return (len * (st + en))/2;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
for(int i=0; i<t; i++) {
long long N, L, S;
cin >> N >> L >> S;
if(S < ap_sum(1, L) || S > ap_sum(N - L + 1, L)) {
cout << "NO\n";
} else {
cout << "YES\n";
}
}
return 0;
}
/*
*/, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code: class Solution {
public static int Rare(int n, int k){
while(n>0){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code: def Rare(N,K):
while N>0:
if(N%10)%K!=0:
return 0
N=N//10
return 1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code:
int Rare(int n, int k){
while(n){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code:
int Rare(int n, int k){
while(n){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: def sample(n):
if n<200:
print(200-n)
elif n<400:
print(400-n)
elif n<500:
print(500-n)
else:
div=n//100
if div*100==n:
print(0)
else:
print((div+1)*100-n)
n=int(input())
sample(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
if(n <= 200){
cout<<200-n;
return;
}
if(n <= 400){
cout<<400-n;
return;
}
int ans = (100-n%100)%100;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans=0;
if(n <= 200){
ans = 200-n;
}
else if(n <= 400){
ans=400-n;
}
else{
ans = (100-n%100)%100;
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
nr = list(map(int,input().split()))
n = nr[0]
r = nr[1]
_ = list(map(int,input().split()))
num1 = _[0]
num2 = _[1]
gcd = find_gcd(num1, num2)
for i in range(2,n):
gcd = find_gcd(gcd, _[i])
if gcd>=r:
print(gcd)
else:
while r:
num = 0
idx = None
for i in range(n):
if _[i]%r != 0:
num += 1
idx = i
if num > 1:
break
if idx!= None:
temp = _[idx]
if num == 1:
_[idx] = r
num1 = _[0]
num2 = _[1]
tgcd = find_gcd(num1, num2)
for i in range(2,n):
tgcd = find_gcd(tgcd, _[i])
gcd = max(tgcd,gcd)
_[idx] = temp
break
r -= 1
print(gcd), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main {
public static long mod = 1000000007;
public static long mod2 = 998244353;
public static void main(String[] args) throws java.lang.Exception {
Reader sc = new Reader();
FastNum in = new FastNum(System.in);
Writer out = new Writer(System.out);
int n = in.nextInt();
int r = in.nextInt();
int[] a = in.nextIntArray(n);
int ans = a[0];
for (int i = 1; i < n; i++) {
ans = gcd(ans, a[i]);
}
int max = MaxGCD(a, n);
for (int i = 1; i <= r; i++) {
ans = Math.max(ans, gcd(max, i));
}
out.println(ans);
out.flush();
}
static int MaxGCD(int a[], int n) {
int[] Prefix = new int[n + 2];
int[] Suffix = new int[n + 2];
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1) {
Prefix[i] = gcd(Prefix[i - 1],
a[i - 1]);
}
Suffix[n] = a[n - 1];
for (int i = n - 1; i >= 1; i -= 1) {
Suffix[i] = gcd(Suffix[i + 1],
a[i - 1]);
}
int ans = Math.max(Suffix[2], Prefix[n - 1]);
for (int i = 2; i < n; i += 1) {
ans = Math.max(ans, gcd(Prefix[i - 1],
Suffix[i + 1]));
}
return ans;
}
static long modSum(long p, long q) {
if (q > 0) {
return (p + q) % mod;
} else {
return (p + q + mod) % mod;
}
}
static long invMod(long p, long q, long m) {
long expo = m - 2;
while (expo != 0) {
if ((expo & 1) == 1) {
p = (p * q) % m;
}
q = (q * q) % m;
expo >>= 1;
}
return p;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long power(long a, long b) {
if (b == 0)
return 1;
long answer = power(a, b / 2) % mod;
answer = (answer * answer) % mod;
if (b % 2 != 0)
answer = (answer * a) % mod;
return answer;
}
public static void swap(int x, int y) {
int t = x;
x = y;
y = t;
}
public static long min(long a, long b) {
if (a < b) return a;
return b;
}
public static long divide(long a, long b) {
return (a % mod * (power(b, mod - 2) % mod)) % mod;
}
public static long nCr(long n, long r) {
long answer = 1;
long k = min(r, n - r);
for (long i = 0; i < k; i++) {
answer = (answer % mod * (n - i) % mod) % mod;
answer = divide(answer, i + 1);
}
return answer % mod;
}
public static boolean plaindrome(String str) {
StringBuilder sb = new StringBuilder();
sb.append(str);
return (str.equals((sb.reverse()).toString()));
}
public static class Pair {
int a;
int b;
Pair(int s, int e) {
a = s;
b = e;
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair pair = (Pair) o;
return a == pair.a && b == pair.b;
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
}
static class Assert {
static void check(boolean e) {
if (!e) {
throw new AssertionError();
}
}
}
static class FastNum implements AutoCloseable {
InputStream is;
byte buffer[] = new byte[1 << 16];
int size = 0;
int pos = 0;
FastNum(InputStream is) {
this.is = is;
}
int nextChar() {
if (pos >= size) {
try {
size = is.read(buffer);
} catch (IOException e) {
throw new IOError(e);
}
pos = 0;
if (size == -1) {
return -1;
}
}
Assert.check(pos < size);
int c = buffer[pos] & 0xFF;
pos++;
return c;
}
int nextInt() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
int n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Integer.MIN_VALUE / 10
|| n == Integer.MIN_VALUE / 10 && d <= -(Integer.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
int n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Integer.MAX_VALUE / 10 || n == Integer.MAX_VALUE / 10 && d <= Integer.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
char nextCh() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
return (char) c;
}
long nextLong() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
long n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Long.MIN_VALUE / 10
|| n == Long.MIN_VALUE / 10 && d <= -(Long.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
long n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Long.MAX_VALUE / 10 || n == Long.MAX_VALUE / 10 && d <= Long.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
char[] nextCharArray(int n) {
char[] arr = new char[n];
for (int i = 0; i < n; i++) {
arr[i] = nextCh();
}
return arr;
}
@Override
public void close() {
}
}
static class Writer extends PrintWriter {
public Writer(java.io.Writer out) {
super(out);
}
public Writer(java.io.Writer out, boolean autoFlush) {
super(out, autoFlush);
}
public Writer(OutputStream out) {
super(out);
}
public Writer(OutputStream out, boolean autoFlush) {
super(out, autoFlush);
}
public void printArray(int[] arr) {
for (int j : arr) {
print(j);
print(' ');
}
println();
}
public void printArray(long[] arr) {
for (long j : arr) {
print(j);
print(' ');
}
println();
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 1e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
vi divisor(N + 1);
readb(n, r);
FORD (i, r, 1)
{
for (int j = i; j <= N; j += i)
if (!divisor[j]) divisor[j] = i;
}
readarr(a, n);
int pre[n + 1] = {}, suf[n + 2] = {};
FOR (i, 1, n) pre[i] = __gcd(pre[i - 1], a[i]);
FORD (i, n, 1) suf[i] = __gcd(suf[i + 1], a[i]);
int ans = pre[n];
FOR (i, 1, n)
{
int x = __gcd(pre[i - 1], suf[i + 1]);
ans = max(ans, divisor[x]);
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: def For_Loop(n):
string = ""
for i in range(1, n+1):
if i % 2 == 0:
string += "%s " % i
return string
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int k = sc.nextInt();
int digit = 1;
int m = 1000000007;;
int arr[] = new int[k+1];
arr[0] = 1;
arr[1] = 1;
for(int i=2; i<=k; i++){
for(int j=i;j>=1;j--){
arr[j] += arr[j-1];
arr[j] = arr[j]%m;
}
}
for(int i=0;i<=k;i++){
System.out.print(arr[i] + " ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: def generateNthRow (N):
prev = 1
print(prev, end = '')
for i in range(1, N + 1):
curr = (prev * (N - i + 1)) // i
print("", curr%1000000007, end = '')
prev = curr
N = int(input())
generateNthRow(N), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
signed main()
{
int t,n;
t=1;
vector<vector<long long> > v(3002);
for(int i=0; i<=3001; i++)
{
for (int j=0; j<=i; j++)
{
if (j==0 || j==i) v[i].push_back(1);
else v[i].push_back((v[i-1][j]%mod + v[i-1][j-1]%mod)%mod);
}
}
while (t--)
{
cin>>n;
n=n+1;
for (int i=0; i<v[n-1].size(); i++) cout<<v[n-1][i]<<" ";
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N , you need to calculate number of distinct values of gcd(i, N) where i is between 1<=i<=1e18.
Note:-You need to check for all the possible values of i.Input contains a single integer N.
Constraints:-
1<=N<=10^10
Print the number of distinct numbers of gcd(i, N) where i is between 1<=i<=1e18.Input:
16
Output:
5
Explanation:-
all disticnt numbers are - 1,2,4,8,16
Input:
3248
Output:
20 , I have written this Solution Code: from math import sqrt
a=int(input())
c=0
i=2
while(i*i < a):
if(a%i==0):
c=c+1
i=i+1
for i in range(int(sqrt(a)), 0, -1):
if (a % i == 0):
c=c+1
print(c+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N , you need to calculate number of distinct values of gcd(i, N) where i is between 1<=i<=1e18.
Note:-You need to check for all the possible values of i.Input contains a single integer N.
Constraints:-
1<=N<=10^10
Print the number of distinct numbers of gcd(i, N) where i is between 1<=i<=1e18.Input:
16
Output:
5
Explanation:-
all disticnt numbers are - 1,2,4,8,16
Input:
3248
Output:
20 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
long N = Long.parseLong(br.readLine());
long count = 0;
for(long i = 1; i*i <= N; i++){
if(N % i == 0){
if(i*i < N){
count += 2;
}else if(i*i == N){
count++;
}
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N , you need to calculate number of distinct values of gcd(i, N) where i is between 1<=i<=1e18.
Note:-You need to check for all the possible values of i.Input contains a single integer N.
Constraints:-
1<=N<=10^10
Print the number of distinct numbers of gcd(i, N) where i is between 1<=i<=1e18.Input:
16
Output:
5
Explanation:-
all disticnt numbers are - 1,2,4,8,16
Input:
3248
Output:
20 , I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
long long n;
cin>>n;
long long x=sqrt(n);
unordered_map<long long ,int> m;
for(int i=1;i<=x;i++){
if(n%i==0){m[i]++;
if(n/i!=i){m[n/i]++;}}
}
cout<<m.size();
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton's garden has N apple trees. Initially, there are A<sub>i</sub> apples attached to the ith tree. Also, there are M apples which newton has to attach to these tress for testing gravity laws.
In one minute, at most B<sub>i</sub> apples fall from the ith tree.
Help newton find the minimum time he has to wait after which he can safely sit under any apple tree.The first line contains two space-separated integers – N and M indicating the number of apple trees and the number of apples to be attached.
Each of the next N lines contains two integers A<sub>i</sub> and B<sub>i</sub> specifying the initial count of apples and rate of fall of apples per minute for ith tree.
<b> Constraints: </b>
1 <= N <= 2*10<sup>5</sup>
0 <= A<sub>i</sub> <= 10<sup>6</sup>
1 <= M, B<sub>i</sub> <= 10<sup>6</sup>Output minimum time Newton has to waitInput:
3 6
4 3
7 4
1 5
Output:
2
Explanation:
Attach 2, 1, 3 apples to 1st, 2nd, 3rd tree respectively. Then, 2, 2, 1 minutes are required for all apples to fall from 1st, 2nd, 3rd tree respectively. So, Wait time is max(2, 2, 1) = 2., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int N = 2*1e5+5;
ll n,m;
vector<ll> a(N),b(N);
bool possible(ll time) {
ll lim = 0;
for(int i=0 ; i<n ; i++) lim += b[i] * time;
ll tot = m;
for(int i=0 ; i<n ; i++) tot += a[i];
if (tot > lim) return false;
for(int i=0 ; i<n ; i++) if (a[i] > b[i]*time) return false;
return true;
}
int main(){
cin >> n >> m;
for(int i=0 ; i<n ; i++){
cin >> a[i] >> b[i];
}
ll lo = 1, hi = 2*1e6+1000;
while(hi - lo > 1) {
ll mid = (hi+lo)/2;
if (possible(mid)){
hi = mid;
}else{
lo = mid;
}
}
cout << hi << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Your task is to implement a stack using an array and perform given queries
<b>Note</b>: Description of each query is given in the <b>input and output format</b>User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>push()</b>:- that takes the integer to be added and the maximum size of the array as a parameter.
<b>pop()</b>:- that takes no parameter.
<b>top()</b> :- that takes no parameter.
Constraints:
1 <= N(number of queries) <= 10<sup>3</sup>During a <b>pop</b> operation if the stack is empty you need to print "<b>Stack underflow</b>",
during <b>push</b> operation, if the maximum size of the array is reached you need to print "<b>Stack overflow</b>", <br> during <b>top</b> operation, you need to print the element which is at the top
if the stack is empty you need to print "<b>Empty stack</b>".
<b>Note</b>:- Each message or element is to be printed on a new line
Sample Input:-
6 3
pop
push 3
push 2
push 4
push 6
top
Sample Output:-
Stack underflow
Stack overflow
4
Explanation:-
Here maximum size of the array is 3, so element 6 can not be added to stack
Sample input:-
8 4
push 2
top
push 4
top
push 6
top
push 8
top
Sample Output:-
2
4
6
8
, I have written this Solution Code: void push(int x,int k)
{
if (top >= k-1) {
System.out.println("Stack overflow");
}
else {
a[++top] = x;
}
}
void pop()
{
if (top < 0) {
System.out.println("Stack underflow");
}
else {
int x = a[top--];
}
}
void top()
{
if (top < 0) {
System.out.println("Empty stack");
}
else {
int x = a[top];
System.out.println(x);
}
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine().trim();
char n[] = s.toCharArray();
int posMin = 0;
for(int i=0;i<n.length;i++){
if(n[posMin]>n[i])
posMin = i;
}
for(int i=0;i<n.length;i++)
n[i] = n[posMin];
System.out.print(String.valueOf(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
char c='z';
for(auto r:s)
c=min(c,r);
for(int i=0;i<s.length();++i)
cout<<c;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: s=input()
l=len(s)
k=min(s)
m=k*l
print(m), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2[] = str1.split(" ");
int n = Integer.parseInt(str2[0]);
int k = Integer.parseInt(str2[1]);
String str3 = br.readLine();
String str4[] = str3.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str4[i]);
}
Arrays.sort(arr);
int i=0,j=2;
long ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((arr[j]-arr[i])>k){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: n, p = input().split(' ')
n = int(n)
p = int(p)
arr = input().split(' ')[:n]
for i in range(n):
arr[i] = int(arr[i])
arr.sort()
i = 0
k = 2
count = 0
while k!=n:
if i == k-1:
k = k + 1
continue
if (arr[k] - arr[i]) <= p:
count = count + (int)(((k-i)*(k-i-1))/2)
k = k + 1
else:
i = i + 1
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,p;
cin>>n>>p;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int i=0,j=2;
int ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((a[j]-a[i])>p){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a stack containing some integers, your task is to reverse the given stack.
Note:- Try to do this question using recursion, do not use any loop.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Reverse_stack()</b> that takes no parameter.
</b>Constraints:</b>
1 <= Elements in stack <= 100
<b> Custom Input: </b>
First line of input should contain the number of elements N of the stack, the next line of input should contain N space separated integers depicting the elements of the stack.
You don't need to return or print anything just complete the given function.
Note:- For the custom input if your code is correct then the elements will be printed in the same order.Sample Input:-
Stack = {1, 2, 3, 4, 5}, where top = 5
Sample Output:-
Stack = {5, 4, 3, 2, 1} where top = 1, I have written this Solution Code: static Stack <Integer> St = new Stack();
static void Reverse_Stack(){
if(St.size()==0){
return;
}
int x=St.peek();
St.pop();
Reverse_Stack();
bottom_insert(x);
}
static void bottom_insert(int x){
if(St.isEmpty()){
St.push(x);
}
else{
int a = St.peek();
St.pop();
bottom_insert(x);
St.push(a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a stack containing some integers, your task is to reverse the given stack.
Note:- Try to do this question using recursion, do not use any loop.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>Reverse_stack()</b> that takes no parameter.
</b>Constraints:</b>
1 <= Elements in stack <= 100
<b> Custom Input: </b>
First line of input should contain the number of elements N of the stack, the next line of input should contain N space separated integers depicting the elements of the stack.
You don't need to return or print anything just complete the given function.
Note:- For the custom input if your code is correct then the elements will be printed in the same order.Sample Input:-
Stack = {1, 2, 3, 4, 5}, where top = 5
Sample Output:-
Stack = {5, 4, 3, 2, 1} where top = 1, I have written this Solution Code: n=int(input())
lst=[int(x) for x in input().split()]
for i in lst:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int n = Integer.parseInt(in.readLine());
int []donationList = new int[n];
int[] defaulterList = new int[n];
long totalDonations = 0L;
StringTokenizer st = new StringTokenizer(in.readLine());
for(int i=0 ; i<n ; i++){
donationList[i] = Integer.parseInt(st.nextToken());
}
int max = Integer.MIN_VALUE;
for(int i=0; i<n; i++){
totalDonations += donationList[i];
max = Math.max(max,donationList[i]);
if(i>0){
if(donationList[i] >= max)
defaulterList[i] = 0;
else
defaulterList[i] = max - donationList[i];
}
totalDonations += defaulterList[i];
}
for(int i=0; i<n ;i++){
System.out.print(defaulterList[i]+" ");
}
System.out.println();
System.out.print(totalDonations);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: n = int(input())
a = input().split()
b = int(a[0])
sum = 0
for i in a:
if int(i)<b:
print(b-int(i),end=' ')
else:
b = int(i)
print(0,end=' ')
sum = sum+b
print()
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After the decimation, the world went into chaos. People had to rebuild the planet so Shield came up with a donation strategy. They feel all the rich guys need to donate more than the poor guys. So, they make a rule. They would make a donation list in which the donation of each person would be shown. But the rule is that a person can’t pay less than what has already been paid before them. Find the extra amount each person will pay, and also, tell shield the total amount of donation.The first line contains n, the total number of people donating. The next line contains n space-separated integers denoting the amount of money paid by the people. The amounts are mentioned in the order in which the people paid.
<b>Constraints:-</b>
1 <= n <= 100000
0 <= money <= 100000The first line contains the extra money that each student has to pay after their teacher applied the rule. The second line contains the total amount collected by the teacher at the end.Sample Input 1:-
10
1 2 3 2 4 3 6 6 7 6
Sample Output 1:-
0 0 0 1 0 1 0 0 0 1
43
Sample Input 2:-
7
10 20 30 40 30 20 10
Sample Output 2:-
0 0 0 0 10 20 30
220, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
void solve(){
int n;
cin>>n;
int a[n];
int ma=0;
int cnt=0;
//map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
ma=max(ma,a[i]);
cout<<ma-a[i]<<" ";
cnt+=ma-a[i];
cnt+=a[i];
//m[a[i]]++;
}
cout<<endl;
cout<<cnt<<endl;
}
signed main(){
int t;
t=1;
while(t--){
solve();}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, find the index which partitions the arrays into 2 subarrays such that absolute difference of their sums is minimized.
If we partition with respect to index i (1<=i<=N) subarrays will be:- (A1,A2,A3.....Ai-1) , (Ai+1,Ai+2,......An)First line of input contains the size of array N, next line contains N space separated integers depicting the values of array.
<b>Constraint</b>
1<=N<=10^5
1<=Arr[i]<=10^7
Print the index for minimum absolute difference. If more than one answer exists print the minimum index.Sample input:-
4
1 2 3 4
Sample output:-
3
Explanation:-
For index 1 absolute difference = 2 + 3 + 4 = 9
For index 2 absolute difference = 3 + 4 - 1 = 6
For index 3 absolute difference = 4 - 1 - 2 = 1
For index 4 absolute difference = 1 + 2 + 3 = 6, I have written this Solution Code: 'use strict'
function Main(input)
{
const inputList = input.split('\n');
const arrSize = Number(inputList[0].split(' ')[0])
const arr = inputList[1].split(' ').
map(x => Number(x))
var res = minAbsDiff(arr, arrSize)
console.log(res)
}
function minAbsDiff(arr, arrSize)
{
var sum = 0
var list = new Array(arrSize)
for(var i = 0; i < arrSize; i++)
{
sum += arr[i]
list[i] = sum
}
var res = 9223372036854775807
var pre = 0
var cnt = 0
for(var i=0;i<arrSize;i++){
if(Math.abs(sum-list[i]-pre)<res)
{
res=Math.abs(sum-list[i]-pre)
cnt=i
}
pre+=arr[i]
}
return (cnt+1)
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, find the index which partitions the arrays into 2 subarrays such that absolute difference of their sums is minimized.
If we partition with respect to index i (1<=i<=N) subarrays will be:- (A1,A2,A3.....Ai-1) , (Ai+1,Ai+2,......An)First line of input contains the size of array N, next line contains N space separated integers depicting the values of array.
<b>Constraint</b>
1<=N<=10^5
1<=Arr[i]<=10^7
Print the index for minimum absolute difference. If more than one answer exists print the minimum index.Sample input:-
4
1 2 3 4
Sample output:-
3
Explanation:-
For index 1 absolute difference = 2 + 3 + 4 = 9
For index 2 absolute difference = 3 + 4 - 1 = 6
For index 3 absolute difference = 4 - 1 - 2 = 1
For index 4 absolute difference = 1 + 2 + 3 = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String str1 = br.readLine();
String[] str2 = str1.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Integer.parseInt(str2[i]);
}
long sum1 = arr[0];
long sum2 = arr[n-1];
int l = 0;
int r = n-1;
while(l < r) {
if(sum1 < sum2) {
++l;
sum1 = sum1 + arr[l];
} else {
--r;
sum2 = sum2 + arr[r];
}
}
System.out.print(r+1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, find the index which partitions the arrays into 2 subarrays such that absolute difference of their sums is minimized.
If we partition with respect to index i (1<=i<=N) subarrays will be:- (A1,A2,A3.....Ai-1) , (Ai+1,Ai+2,......An)First line of input contains the size of array N, next line contains N space separated integers depicting the values of array.
<b>Constraint</b>
1<=N<=10^5
1<=Arr[i]<=10^7
Print the index for minimum absolute difference. If more than one answer exists print the minimum index.Sample input:-
4
1 2 3 4
Sample output:-
3
Explanation:-
For index 1 absolute difference = 2 + 3 + 4 = 9
For index 2 absolute difference = 3 + 4 - 1 = 6
For index 3 absolute difference = 4 - 1 - 2 = 1
For index 4 absolute difference = 1 + 2 + 3 = 6, I have written this Solution Code: import math
n = int(input())
lst = list(map(int, input().strip().split()))
cuLstL = []
cuLstR = []
summ = 0
for i in range(len(lst)):
summ += lst[i]
cuLstL.append(summ)
summ = 0
for i in range(len(lst)-1, -1, -1):
summ += lst[i]
cuLstR.append(summ)
cuLstR = cuLstR[::-1]
l = []
for i in range(n):
l.append(abs(cuLstL[i]-cuLstR[i]))
print(l.index(min(l))+1)
'''
for i in range(len(lst)):
res = max(res, cuLst[i] - sink)
sink = min(sink, cuLst[i])
print(res,sink)
print(res)
''', In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code:
void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: public static void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: function patternMaking(N)
{
for(let j=1; j <= 2*N - 1; j++) {
let k;
if(j<= N) {
k = j;
} else {
k = 2*N - j;
}
let res = "";
for(let i=1; i<=2*k-1; i++) {
if(i<=k) {
res += i + " ";
} else {
res += 2*k - i + " ";
}
}
console.log(res);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: def pattern_making(n):
for i in range(1, n+1):
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=n-1
while i>=1 :
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=i-1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.StringTokenizer;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
};
public static void main(String[] args) {
FastReader sc = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextLong();
}
Arrays.sort(arr);
long min=Long.MAX_VALUE;
long max=Long.MIN_VALUE;
for(int i=0;i<n/2;i++)
{
min=Math.min(min,arr[n-i-1]+arr[i]);
max=Math.max(max,arr[n-i-1]+arr[i]);
}
System.out.println(max-min);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: n = int(input())
bruh = input().strip().split(" ")
for i in range(len(bruh)):
bruh[i] = int(bruh[i])
bruh.sort()
maxSum = 0
minSum = 100000000000000000000
for i in range(len(bruh)//2):
add = bruh[i] + bruh[-(i+1)]
if add > maxSum:
maxSum = add
if add < minSum:
minSum = add
print(maxSum - minSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pree says he stole Daenerys' dragons, and would return them to Daenerys only if she can pair the dragons in a peculiar way.
There are N (N is even) dragons, the strength of i<sup>th</sup> dragon is A<sub>i</sub>. She needs to create N/2 pairs of the dragons. The strength of the dragon pair is the sum of the strengths of dragons in the pair. She needs to pair them in such a way that the difference between the strength of dragon pair with maximum strength and the strength of the dragon pair with minimum strength is minimised.
Now, you need to find the minimum difference in strengths if she has paired them correctly.The first line of the input contains an integer N, the number of dragons.
The second line of the input contains N integers, the strengths of the dragons.
Constraints
1 <= N <= 200000 (so many dragons, huh)
1 <= A<sub>i</sub> <= 10<sup>9</sup> for all values of i
N is divisible by 2Output a single integer, the answer to the problem.Sample Input
4
1 2 1 2
Sample Output
0
Explanation
Daenerys pairs 1st and the 2nd dragons together; the 3rd and the 4th dragon together
Sample Input
6
2 3 2 4 5 1
Sample Output
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> v(n);
For(i, 0, n){
cin>>v[i];
}
sort(all(v));
vector<int> cur;
For(i, 0, n/2){
cur.pb(v[i]+v[n-i-1]);
}
sort(all(cur));
cout<<cur[n/2-1]-cur[0];
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function insertionSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void insertionSort(int[] arr){
for(int i = 0; i < arr.length-1; i++){
for(int j = i+1; j < arr.length; j++){
if(arr[i] > arr[j]){
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
}
}
}
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
while(T > 0){
int n = scan.nextInt();
int arr[] = new int[n];
for(int i = 0; i<n; i++){
arr[i] = scan.nextInt();
}
insertionSort(arr);
for(int i = 0; i<n; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
T--;
System.gc();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: def InsertionSort(arr):
arr.sort()
return arr, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. In one move, you can swap any two elements of this array. Find out whether you can make this array strictly increasing or not after any (possibly 0) number of moves.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>6</sup>Print "YES", if you can make the array strictly increasing, else print "NO", without the quotes.Sample Input:
5
5 9 4 5 3
Sample Output:
NO
Explaination:
After any number of moves, you cannot make this array strictly increasing., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(a.begin(), a.end());
for(int i = 1; i < n; i++){
if(a[i] == a[i - 1]){
cout << "NO";
return 0;
}
}
cout << "YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: import java.util.InputMismatchException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
int MAX = (int) 1e5, MOD = (int)1e9+7;
void solve(int TC) {
long n = nl();
long k = nl();
int p = 0;
while(n>0 && n%2==0) {
n/=2;
++p;
}
if(p>=k) {pn(0);return;}
k -= p;
long ans = (k+3L)/4L;
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
long pow(long a, long b) {
if(b==0 || a==1) return 1;
long o = 1;
for(long p = b; p > 0; p>>=1) {
if((p&1)==1) o = (o*a) % MOD;
a = (a*a) % MOD;
} return o;
}
long inv(long x) {
long o = 1;
for(long p = MOD-2; p > 0; p>>=1) {
if((p&1)==1)o = (o*x)%MOD;
x = (x*x)%MOD;
} return o;
}
long gcd(long a, long b) { return (b==0) ? a : gcd(b,a%b); }
int gcd(int a, int b) { return (b==0) ? a : gcd(b,a%b); }
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
double PI = 3.141592653589793238462643383279502884197169399;
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: [n,k]=[int(j) for j in input().split()]
a=0
while n%2==0:
a+=1
n=n//2
if k>a:
print((k-a-1)//4+1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,k;
cin>>n>>k;
while(k&&n%2==0){
n/=2;
--k;
}
cout<<(k+3)/4;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton has N baskets containing apples. There are A<sub>i</sub> apples in ith basket. Newton wants to gift one apple to each of his M friends.
To do that he will select some baskets and gift all apples in these baskets to his friends. <b>Note</b> that he will gift apples if and only if there are exactly M apples in these subset of baskets.
Help him determine if it is possible to gift apple or not.The first line contains two space-separated integers – N and M.
The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub>.
<b> Constraints: </b>
1 ≤ N ≤ 100
1 ≤ M ≤ 10<sup>11</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>
1 <= ∏ A<sub>i</sub> <= 10<sup>100</sup>Output "Yes" (without quotes) if Newton can distribute apples among his friends else "No" (without quotes).INPUT:
4 8
1 3 3 4
OUTPUT:
Yes
EXPLANATION:
Newton can use 1st, 2nd, 4th basket to give one apple to all his friends., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const ll b = 10000;
ll pre[1000000];
int main(){
ll n,x;
cin >> n >> x;
vector<ll> a(n);
for(ll i=0 ; i<n ; i++){
cin >> a[i];
}
fill(pre, pre + 1000000, -1);
pre[0] = 0;
vector<pair<ll,ll>> v;
for(ll i=0 ; i<n ; i++){
if (a[i]>=b) v.push_back({ a[i],i });
else{
for(ll j=1000000-1 ; j>=0 ; j--) {
if (pre[j] < 0) continue;
if (j + a[i] < 1000000 && pre[j+a[i]] < 0) {
pre[j + a[i]] = j;
}
}
}
}
ll len = v.size();
for(ll i=0 ; i<(1 << len) ; i++){
ll sum = 0;
for(ll j=0 ; j<len ; j++) {
if (i & (1 << j)) sum += v[j].first;
}
ll d = x - sum;
if (d >= 0 && d < 1000000) {
if (pre[d] >= 0) {
cout << "Yes\n";
return 0;
}
}
}
cout << "No" << "\n";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
Output
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int n = s.length();
int left = 0, right = 0;
int maxlength = 0;
for(int i = 0; i < n; i++)
{
if (s.charAt(i) == '(')
left++;
else
right++;
if (left == right)
maxlength = Math.max(maxlength, 2 * right);
else if (right > left)
left = right = 0;
}
left = right = 0;
for(int i = n - 1; i >= 0; i--)
{
if (s.charAt(i) == '(')
left++;
else
right++;
if (left == right)
maxlength = Math.max(maxlength, 2 * left);
else if (left > right)
left = right = 0;
}
System.out.println(maxlength);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
Output
4, I have written this Solution Code: def longestValidParentheses( S):
stack, ans = [-1], 0
for i in range(len(S)):
if S[i] == '(': stack.append(i)
elif len(stack) == 1: stack[0] = i
else:
stack.pop()
ans = max(ans, i - stack[-1])
return ans
n=input()
print(longestValidParentheses(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of opening and closing parenthesis '(' and ')'. Find length of the longest valid parenthesis substring.Each test case have one line string S of character '(' and ')' of length N.
Constraints:
1 <= N <= 10^5Print the length of the longest valid parenthesis substring.Input
((()
Output
2
Input
)()())
Output
4, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
string s;
cin >> s;
int n = s.length();
s = "#" + s;
stack<int> st;
st.push(0);
int ans = 0;
for(int i = 1; i <= n; i++){
if(s[i] == '(')
st.push(i);
else{
if(s[st.top()] == '('){
int x = st.top();
a[i] = i-x+1 + a[x-1];
ans = max(ans, a[i]);
st.pop();
}
else
st.push(i);
}
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Consider the integer sequence A = {1, 2, 3, ...., N} i.e. the first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing <b>at most one</b> element from A.
A <b>subarray</b> of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains three integers: N, L, and S.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>9</sup>
1 <= L <= N-1
1 <= S <= 10<sup>18</sup> <b> (Note that S will not fit in a 32-bit integer) </b>For each testcase, print "YES" (without quotes) if a required subarray can exist, and "NO" (without quotes) otherwise.Sample Input:
3
5 3 11
5 3 5
5 3 6
Sample Output:
YES
NO
YES
Sample Explanation:
For the first test case, we can remove 3 from A to obtain A = {1, 2, 4, 5} where {2, 4, 5} is a required subarray of size 3 and sum 11., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long INF = 1e18;
const int INFINT = INT_MAX/2;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x, y) freopen(x, "r", stdin); freopen(y, "w", stdout);
#define out(x) cout << ((x) ? "YES\n" : "NO\n")
#define CASE(x, y) cout << "Case #" << x << ":" << " \n"[y]
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now(); auto end_time = start_time;
#define measure() end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
#define reset_clock() start_time = std::chrono::high_resolution_clock::now();
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll norm(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; g--; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n): adj(n+1) {}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
struct LCA
{
vll tin, tout, level;
vector<vll> up;
ll timer;
void _dfs(vector<vector<int>> &adj, ll x, ll par=-1)
{
up[x][0] = par;
REP(i, 1, 20)
{
if(up[x][i-1] == -1) break;
up[x][i] = up[up[x][i-1]][i-1];
}
tin[x] = ++timer;
for(auto &p: adj[x])
{
if(p != par)
{
level[p] = level[x]+1; _dfs(adj, p, x);
}
}
tout[x] = ++timer;
}
LCA(Graph &G, ll root)
{
int n = G.adj.size();
tin.resize(n); tout.resize(n); up.resize(n, vll(20, -1)); level.resize(n, 0);
timer = 0;
//does not handle forests, easy to modify to handle
_dfs(G.adj, root);
}
ll find_kth(ll x, ll k)
{
ll cur = x;
REP(i, 0, 20)
{
if((k>>i)&1) cur = up[cur][i];
if(cur == -1) break;
}
return cur;
}
bool is_ancestor(ll x, ll y)
{
return tin[x]<=tin[y]&&tout[x]>=tout[y];
}
ll find_lca(ll x, ll y)
{
if(is_ancestor(x, y)) return x;
if(is_ancestor(y, x)) return y;
ll best = x;
REPd(i, 19, 0)
{
if(up[x][i] == -1) continue;
else if(is_ancestor(up[x][i], y)) best = up[x][i];
else x = up[x][i];
}
return best;
}
ll dist(ll a, ll b)
{
return level[a] + level[b] - 2*level[find_lca(a, b)];
}
};
long long ap_sum(long long st, long long len) {
//diff is always 1
long long en = st + len - 1;
return (len * (st + en))/2;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
for(int i=0; i<t; i++) {
long long N, L, S;
cin >> N >> L >> S;
if(S < ap_sum(1, L) || S > ap_sum(N - L + 1, L)) {
cout << "NO\n";
} else {
cout << "YES\n";
}
}
return 0;
}
/*
*/, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code:
void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: public static void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: function patternMaking(N)
{
for(let j=1; j <= 2*N - 1; j++) {
let k;
if(j<= N) {
k = j;
} else {
k = 2*N - j;
}
let res = "";
for(let i=1; i<=2*k-1; i++) {
if(i<=k) {
res += i + " ";
} else {
res += 2*k - i + " ";
}
}
console.log(res);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: def pattern_making(n):
for i in range(1, n+1):
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=n-1
while i>=1 :
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=i-1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let us start our journey by creating a 'PROFILE' table for storing the data of the users of our platform. Create a table profile with 3 fields ( USERNAME VARCHAR (24), FULL_NAME VARCHAR (72), HEADLINE VARCHAR (72) ).
<schema>[{'name': 'PROFILE', 'columns': [{'name': 'USERNAME', 'type': 'VARCHAR (24)'}, {'name': 'FULL_NAME', 'type': 'VARCHAR (72)'}, {'name': 'HEADLINE', 'type': 'VARCHAR (72)'}]}]</schema>NA - User input does not work for this question.The output will be generated in a 2-D array format consisting of rows and columns.nan, I have written this Solution Code: CREATE TABLE PROFILE (
USERNAME VARCHAR(24),
FULL_NAME VARCHAR(72),
HEADLINE VARCHAR(72)
);
, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Find whether the given number is a perfect square.
Note: Don't use the C++ library function sqrt().
Expected Time Complexity: O(log n)First line contains an integer N
Constraints
1 <= N <= 100000Print "YES" if x is a perfect square otherwise print "NO".Sample Input 1:
16
Output:
YES
Sample Input 2:
7
Output:
NO, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main
{
public static void main(String[] args) throws IOException
{
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(reader.readLine());
int sqrt = (int) Math.sqrt(n);
if(sqrt*sqrt == n)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Find whether the given number is a perfect square.
Note: Don't use the C++ library function sqrt().
Expected Time Complexity: O(log n)First line contains an integer N
Constraints
1 <= N <= 100000Print "YES" if x is a perfect square otherwise print "NO".Sample Input 1:
16
Output:
YES
Sample Input 2:
7
Output:
NO, I have written this Solution Code: x = int(input())
def is_perfect_square(n):
x = n // 2
y = set([x])
while x * x != n:
x = (x + (n // x)) // 2
if x in y: return False
y.add(x)
return True
if(is_perfect_square(x)):
print("YES")
else:
print("NO")
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Find whether the given number is a perfect square.
Note: Don't use the C++ library function sqrt().
Expected Time Complexity: O(log n)First line contains an integer N
Constraints
1 <= N <= 100000Print "YES" if x is a perfect square otherwise print "NO".Sample Input 1:
16
Output:
YES
Sample Input 2:
7
Output:
NO, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long int
#define f(n) for(int i=0;i<n;i++)
#define fo(n) for(int j=0;j<n;j++)
#define foo(n) for(int i=1;i<=n;i++)
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define vp vector<pii>
#define test int tt; cin>>tt; while(tt--)
#define mod 1000000007
void fastio()
{
ios_base::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("Input 1.txt", "r", stdin);
freopen("Output 1.txt", "w", stdout);
#endif
}
bool check(ll n) {
ll l = 0, r = n;
ll ans = -1;
while (l <= r) {
ll a = (2 * l + r) / 3;
ll b = (l + 2 * r) / 3;
bool x = (a * a >= n);
bool y = (b * b >= n);
if (x == false && y == false) {
l = b + 1;
}
else if ((x == true) && (y == true)) {
ans = a;
r = a - 1;
}
else {
ans = b;
l = a + 1;
r = b - 1;
}
}
return (ans * ans == n);
}
int main() {
fastio();
ll n;
cin >> n;
if (check(n)) {
cout << "YES";
}
else {
cout << "NO";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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