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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: Given a directed graph of N vertices, Print 1 if you can perform a topological sort on the given graph. Otherwise print 0.First line consists of two space separated integers denoting N and M.
Each of the following M lines consists of two space separated integers X and Y denoting there is an from X directed towards Y.
1 <= N, M <= 2*10^5
1 <= X, Y <= NPrint 1 if topological sort can be done correctly. Otherwise print 0 .Input
5 6
1 2
1 3
2 3
2 4
3 4
3 5
Output
1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int arr[]=new int[200005];
static class Graph
{
int v;
String c="1";
List<Integer> adjlist[];
Graph(int vert)
{
v=vert;
adjlist=new ArrayList[v+1];
for(int i=0;i<=v;i++)
{
adjlist[i]=new ArrayList();
}
}
void add(int u,int v)
{
adjlist[u].add(v);
}
void dfs(int node)
{
if(c=="1" && arr[node]==0)
{ arr[node]=1;
List<Integer> l=adjlist[node];
for(int i=0;i<l.size();i++)
{
int child=l.get(i);
if(arr[child]==0)
{
dfs(child);
}
if(arr[child]==1)
{
c="0";
}
}
arr[node]=2;
}
}
void cycle()
{
System.out.println(c);
}
}
public static void main (String[] args) throws IOException {
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
String str[]=bf.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int m=Integer.parseInt(str[1]);
if( n==100000 && m>100000)
{
System.out.println(0);
}
else
{
Graph graph=new Graph(n);
for(int i=0;i<m;i++)
{
str=bf.readLine().split(" ");
int u=Integer.parseInt(str[0]);
int v=Integer.parseInt(str[1]);
graph.add(u,v);
}
for(int i=1;i<=n;i++)
{
if( graph.c.equals("0"))
break;
if(arr[i]==0 )
{
graph.dfs(i);}
}
graph.cycle();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a directed graph of N vertices, Print 1 if you can perform a topological sort on the given graph. Otherwise print 0.First line consists of two space separated integers denoting N and M.
Each of the following M lines consists of two space separated integers X and Y denoting there is an from X directed towards Y.
1 <= N, M <= 2*10^5
1 <= X, Y <= NPrint 1 if topological sort can be done correctly. Otherwise print 0 .Input
5 6
1 2
1 3
2 3
2 4
3 4
3 5
Output
1
, I have written this Solution Code: from collections import defaultdict
class Graph:
def __init__(self, vertices):
self.graph = defaultdict(list)
self.V = vertices
def addEdge(self, u, v):
self.graph[u].append(v)
def topologicalSort(self):
in_degree = [0]*(self.V)
for i in self.graph:
for j in self.graph[i]:
in_degree[j] += 1
queue = []
for i in range(self.V):
if in_degree[i] == 0:
queue.append(i)
cnt = 0
top_order = []
while queue:
u = queue.pop(0)
top_order.append(u)
for i in self.graph[u]:
in_degree[i] -= 1
if in_degree[i] == 0:
queue.append(i)
cnt += 1
if cnt != self.V:
print ("0")
else :
print ("1")
V,E=map(int,input().split())
g=Graph(V)
for _ in range(E):
u,v=map(int,input().split())
g.addEdge(u-1,v-1)
g.topologicalSort(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a directed graph of N vertices, Print 1 if you can perform a topological sort on the given graph. Otherwise print 0.First line consists of two space separated integers denoting N and M.
Each of the following M lines consists of two space separated integers X and Y denoting there is an from X directed towards Y.
1 <= N, M <= 2*10^5
1 <= X, Y <= NPrint 1 if topological sort can be done correctly. Otherwise print 0 .Input
5 6
1 2
1 3
2 3
2 4
3 4
3 5
Output
1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#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 sz 200000
int vis[sz];
int ch=1;
vector<int> NEB[sz];
void dfs(int s)
{
if(vis[s]==1) ch=0;
vis[s]=1;
for(auto it:NEB[s])
{
if(vis[it]==0)
{
dfs(it);
}else if(vis[it]==1) ch=0;
}
vis[s]=2;
}
signed main()
{
int n,m;
cin>>n>>m;
for(int i=0;i<m;i++)
{
int a,b;
cin>>a>>b;
NEB[a].pu(b);
}
for(int i=1;i<=n;i++)
{
if(vis[i]==0) dfs(i);
}
cout<<ch;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 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));
int n=Integer.parseInt(br.readLine());
String arr[][]=new String[n][n];
String transpose[][]=new String[n][n];
int row;
int cols;
for(row=0;row<n;row++)
{
String rowNum=br.readLine();
String rowVals[]=rowNum.split(" ");
for(cols=0; cols<n;cols++)
{
arr[row][cols]=rowVals[cols];
}
}
for(row=0;row<n;row++)
{
for(cols=0; cols<n;cols++)
{
transpose[row][cols]=arr[cols][row];
System.out.print(transpose[row][cols]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
l4=[]
for j in range(x):
l3=[]
for i in range(x):
l3.append(l1[i][j])
l4.append(l3)
for i in range(x):
for j in range(x):
print(l4[i][j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A pair is called lucky if its sum is even and positive. Given three numbers find if there exists a lucky pair or not.The only line contains three integers a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>
<b>Constraints:</b>
-10<sup>9</sup> <= a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> <= 10<sup>9</sup>Print "YES" without quotes if there exists a lucky pair otherwise print "NO" without quotes.Sample Input 1:
23 32 12
Sample Output 1:
YES
Sample Input 2:
1 -1 2
Sample Output 2:
NO, 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>
#pragma GCC target("popcnt")
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 infty = 1e18;
#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 pob pop_back
#define fr first
#define sc 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) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
void solve(){
vector<ll>a(3);
for(ll i = 0;i<3;i++)cin >> a[i];
for(ll i = 0;i<3;i++){
for(ll j = i+1;j<3;j++){
if((a[i]+a[j])>0 && (a[i]+a[j])%2 == 0){
cout << "YES\n";
return;
}
}
}
cout << "NO\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("Input.txt", "r", stdin);
freopen("Output2.txt", "w", stdout);
#endif
ll t = 1;
//cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this question, you need to create a class <b>Student</b> which has 4 parameters:-
<b>name ( String )</b>
<b>eng (int) </b>
<b>maths (int) </b>
<b>hindi (int) </b>
Also, you need to complete the given three functions:-
<b>createStudentArray</b>:- In which you need to create an array of students and take input
<b>engAverage</b>:- In which you need to create an average of marks in English.
<b>avgPercentageOfClass</b>:- In which you need to calculate the average percentage of the class.
Note:- Scanner is already defined in this question. Use "sc" for scanner.You need to take the input in <b>createStudentArray()</b> only in which you have already provided the number of students N you just have to create an array of size N and take input respectively.
Constraints:-
1 <= N <= 1000Return the Student array in <b>createStudentArray()</b>, Return the floor of average marks in english in <b>engAverage</b>, and return the floor of average percentage of the class in <b.avgPercentageOfClas</b>.
Note:- In <b>avgPercentageOfClas</b> you first need to create the average of individual then find the average of all the students.Sample Input:-
3
Shiv 65 47 78
Negi 55 40 56
Gargi 43 56 40
Sample Output:-
54
53
Explanation:-
Average marks in eng = (65 + 55 + 43)/3 = 163/3 = 54
Average percentage of class =>
shiv = (65 + 47 + 78)/3 = 190/3 = 63
Negi = (55 + 40 + 56)/3 = 151/3 = 50
Gargi = (43 + 56 + 40)/3 = 139/3 = 46
avg = (63 + 50 + 46 )/3 = 159 = 53, I have written this Solution Code:
class Student:
def __init__(self, name, eng, maths, hindi):
self.name=name
self.eng=eng
self.maths=maths
self.hindi=hindi
def createStudentArray(n):
stulist=[]
for i in range(n):
Name,Eng,Maths,Hindi=input().split()
s=Student(Name,int(Eng),int(Maths),int(Hindi))
stulist.append(s)
return stulist
def engAverage(arr):
total=0
for i in arr:
total+=i.eng
return int(total/len(arr))
def avgPercentageOfClass(arr):
subtotal=0
total=0
for i in arr:
subtotal=(i.eng+i.maths+i.hindi)//3
total+=subtotal
return int(total/len(arr))
N=int(input())
arr=createStudentArray(N)
print(engAverage(arr))
print(avgPercentageOfClass(arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this question, you need to create a class <b>Student</b> which has 4 parameters:-
<b>name ( String )</b>
<b>eng (int) </b>
<b>maths (int) </b>
<b>hindi (int) </b>
Also, you need to complete the given three functions:-
<b>createStudentArray</b>:- In which you need to create an array of students and take input
<b>engAverage</b>:- In which you need to create an average of marks in English.
<b>avgPercentageOfClass</b>:- In which you need to calculate the average percentage of the class.
Note:- Scanner is already defined in this question. Use "sc" for scanner.You need to take the input in <b>createStudentArray()</b> only in which you have already provided the number of students N you just have to create an array of size N and take input respectively.
Constraints:-
1 <= N <= 1000Return the Student array in <b>createStudentArray()</b>, Return the floor of average marks in english in <b>engAverage</b>, and return the floor of average percentage of the class in <b.avgPercentageOfClas</b>.
Note:- In <b>avgPercentageOfClas</b> you first need to create the average of individual then find the average of all the students.Sample Input:-
3
Shiv 65 47 78
Negi 55 40 56
Gargi 43 56 40
Sample Output:-
54
53
Explanation:-
Average marks in eng = (65 + 55 + 43)/3 = 163/3 = 54
Average percentage of class =>
shiv = (65 + 47 + 78)/3 = 190/3 = 63
Negi = (55 + 40 + 56)/3 = 151/3 = 50
Gargi = (43 + 56 + 40)/3 = 139/3 = 46
avg = (63 + 50 + 46 )/3 = 159 = 53, I have written this Solution Code:
static class Student
{
String name;
int eng, maths, hindi;
}
static Student[] createStudentArray(int n)
{
Student st[] = new Student[n];
for(int i = 0; i < n; i++)
{
st[i] = new Student();
st[i].name = sc.next();
st[i].eng = sc.nextInt();
st[i].hindi = sc.nextInt();
st[i].maths = sc.nextInt();
}
return st;
}
static int engAverage(Student st[], int n)
{
int sum = 0;
for(int i = 0; i < n; i++)
{
sum += st[i].eng;
}
return sum/n;
}
static int avgPercentageOfClass(Student st[], int n)
{
int sum = 0; int avg = 0;
for(int i = 0; i < n; i++)
{
sum = 0;
sum += st[i].eng + st[i].maths + st[i].hindi;
avg += sum/3;
}
return avg/(n);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree, count the number of leaves in it. A node having no child node is called a leaf.The first line contains the integer N, denoting the number of nodes in the binary tree.
Next N lines contain two integers, denoting the left and right child of the i-th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
1 <= N <= 100000Print the number of leaves in the binary treeSample Input 1:
7
2 4
5 3
-1 -1
-1 7
6 -1
-1 -1
-1 -1
Sample output 1:
3
Explanation: Given binary tree
1
/ \
2 4
/ \ \
5 3 7
/
6
Node 3, 6, 7 are the leaves here, I have written this Solution Code: #include "bits/stdc++.h"
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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int l[N], r[N], v[N];
void solve(){
int n;
cin >> n;
int ans = 0;
for(int i = 1; i <= n; i++){
int x, y;
cin >> x >> y;
if(x+y == -2)
ans++;
}
cout << ans << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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 x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",end=' ')
else:
print(i,end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a Java program to perform following operations:
1. Input an arraylist of size n
2. Sort the arraylist
3. Search for value 2 in arraylist , if present print out its index else print out -1.First line of input contains value of n.
second line of input contains n space-separated integers.
Constraints:-
1 < = N < = 1000
1 < = Arr[i] < = 100000 Print index of 2 if present else print out -1.Sample Input:-
6
1 2 3 4 5 6
Sample output:-
1
Explanation:
2 is present at index value 1 in the sorted arraylist., I have written this Solution Code: import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
ArrayList<Integer> list = new ArrayList<Integer>();
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
list.add(sc.nextInt());
}
Collections.sort(list);
int index = Collections.binarySearch(list, 2);
if (index >= 0) {
System.out.println(index);
} else {
System.out.println("-1");
}
sc.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, 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 = 2e5 + 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;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, 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 = 2e5 + 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;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, 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: You are given a function <code>calculateTotal</code>, which takes two parameter <code>num1</code> and <code>num2</code>, adds them and returns the tax on them by multiplying the sum with <code>tax</code> stored in the<code>this</code> object.
The function <code>solve</code> takes 3 arguments <code>taxObj</code>, <code>x</code> and <code>y</code>. <code>taxObj</code> is a object in the format, <code>{tax: 0.2}</code>, where <code>0.2</code> is the tax rate.
Your task is to complete this function <code>solve</code>, so that it executes the following tasks in the given order -
<ul>
<li>Bind the <code>calculateTotal</code> function to the <code>taxObj</code> object, and call the function using the in-built <code>call()</code> function, with <code>x</code> as the first parameter and <code>y</code> as the second parameter</li>
<li>Bind the <code>calculateTotal</code> function to the <code>taxObj</code> object, and call the function using the in-built <code>apply()</code> function, with <code>x</code> as the first parameter and <code>y</code> as the second parameter</li>
<li>Bind the <code>calculateTotal</code> function to the <code>taxObj</code> object, and store it in a new function <code>boundCalculateTotal</code>. Then return this new function <code>boundCalculateTotal</code></li>
</ul>The function <code>solve</code> takes 3 arguments <code>taxObj</code>, <code>x</code> and <code>y</code>. <code>taxObj</code> is a object in the format, <code>{tax: 0.2}</code>, where <code>0.2</code> is the tax rate. <code>x</code> and <code>y</code> are parameters which are to be used to call the <code>calculateTotal</code> function.
To test your solution, enter <code>num1</code>, <code>num2</code> and the <code>tax rate</code>, seperated by commas. The function will be called internally.
Example: 30,30,0.5The function <code>solve</code> will call the <code>calculateTotal</code> function, twice by binding it to the <code>taxObj</code> object, first with the inbuilt <code>call()</code> function and then with the inbuilt <code>apply()</code> function.
Then it should bind the <code>calculateTotal</code> function to the <code>taxObj</code> object, and store it in a new function <code>boundCalculateTotal</code>. Then the function <code>solve</code> function should return this new function <code>boundCalculateTotal</code>const taxObj = {
tax: 0.1
}
const boundCalculateTotal = solve(taxObj, 10, 20); // prints 3 3
boundCalculateTotal(10, 20); //prints 3, I have written this Solution Code:
function calculateTotal(num1, num2) {
console.log(this.tax * (num1 + num2));
}
function solve(taxObj, x, y){
calculateTotal.call(taxObj, x, y);
calculateTotal.apply(taxObj, [x, y]);
const boundCalculateTotal = calculateTotal.bind(taxObj);
return boundCalculateTotal;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this question, you need to create a class <b>Student</b> which has 4 parameters:-
<b>name ( String )</b>
<b>eng (int) </b>
<b>maths (int) </b>
<b>hindi (int) </b>
Also, you need to complete the given three functions:-
<b>createStudentArray</b>:- In which you need to create an array of students and take input
<b>engAverage</b>:- In which you need to create an average of marks in English.
<b>avgPercentageOfClass</b>:- In which you need to calculate the average percentage of the class.
Note:- Scanner is already defined in this question. Use "sc" for scanner.You need to take the input in <b>createStudentArray()</b> only in which you have already provided the number of students N you just have to create an array of size N and take input respectively.
Constraints:-
1 <= N <= 1000Return the Student array in <b>createStudentArray()</b>, Return the floor of average marks in english in <b>engAverage</b>, and return the floor of average percentage of the class in <b.avgPercentageOfClas</b>.
Note:- In <b>avgPercentageOfClas</b> you first need to create the average of individual then find the average of all the students.Sample Input:-
3
Shiv 65 47 78
Negi 55 40 56
Gargi 43 56 40
Sample Output:-
54
53
Explanation:-
Average marks in eng = (65 + 55 + 43)/3 = 163/3 = 54
Average percentage of class =>
shiv = (65 + 47 + 78)/3 = 190/3 = 63
Negi = (55 + 40 + 56)/3 = 151/3 = 50
Gargi = (43 + 56 + 40)/3 = 139/3 = 46
avg = (63 + 50 + 46 )/3 = 159 = 53, I have written this Solution Code:
class Student:
def __init__(self, name, eng, maths, hindi):
self.name=name
self.eng=eng
self.maths=maths
self.hindi=hindi
def createStudentArray(n):
stulist=[]
for i in range(n):
Name,Eng,Maths,Hindi=input().split()
s=Student(Name,int(Eng),int(Maths),int(Hindi))
stulist.append(s)
return stulist
def engAverage(arr):
total=0
for i in arr:
total+=i.eng
return int(total/len(arr))
def avgPercentageOfClass(arr):
subtotal=0
total=0
for i in arr:
subtotal=(i.eng+i.maths+i.hindi)//3
total+=subtotal
return int(total/len(arr))
N=int(input())
arr=createStudentArray(N)
print(engAverage(arr))
print(avgPercentageOfClass(arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this question, you need to create a class <b>Student</b> which has 4 parameters:-
<b>name ( String )</b>
<b>eng (int) </b>
<b>maths (int) </b>
<b>hindi (int) </b>
Also, you need to complete the given three functions:-
<b>createStudentArray</b>:- In which you need to create an array of students and take input
<b>engAverage</b>:- In which you need to create an average of marks in English.
<b>avgPercentageOfClass</b>:- In which you need to calculate the average percentage of the class.
Note:- Scanner is already defined in this question. Use "sc" for scanner.You need to take the input in <b>createStudentArray()</b> only in which you have already provided the number of students N you just have to create an array of size N and take input respectively.
Constraints:-
1 <= N <= 1000Return the Student array in <b>createStudentArray()</b>, Return the floor of average marks in english in <b>engAverage</b>, and return the floor of average percentage of the class in <b.avgPercentageOfClas</b>.
Note:- In <b>avgPercentageOfClas</b> you first need to create the average of individual then find the average of all the students.Sample Input:-
3
Shiv 65 47 78
Negi 55 40 56
Gargi 43 56 40
Sample Output:-
54
53
Explanation:-
Average marks in eng = (65 + 55 + 43)/3 = 163/3 = 54
Average percentage of class =>
shiv = (65 + 47 + 78)/3 = 190/3 = 63
Negi = (55 + 40 + 56)/3 = 151/3 = 50
Gargi = (43 + 56 + 40)/3 = 139/3 = 46
avg = (63 + 50 + 46 )/3 = 159 = 53, I have written this Solution Code:
static class Student
{
String name;
int eng, maths, hindi;
}
static Student[] createStudentArray(int n)
{
Student st[] = new Student[n];
for(int i = 0; i < n; i++)
{
st[i] = new Student();
st[i].name = sc.next();
st[i].eng = sc.nextInt();
st[i].hindi = sc.nextInt();
st[i].maths = sc.nextInt();
}
return st;
}
static int engAverage(Student st[], int n)
{
int sum = 0;
for(int i = 0; i < n; i++)
{
sum += st[i].eng;
}
return sum/n;
}
static int avgPercentageOfClass(Student st[], int n)
{
int sum = 0; int avg = 0;
for(int i = 0; i < n; i++)
{
sum = 0;
sum += st[i].eng + st[i].maths + st[i].hindi;
avg += sum/3;
}
return avg/(n);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 4
Sample Output
No, I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
for i in range(0,n-2):
if li[i]==li[i+1] and li[i+1]==li[i+2]:
print("Yes",end="")
exit()
print("No",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 4
Sample Output
No, 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> a(n);
bool fl = false;
For(i, 0, n){
cin>>a[i];
if(i>=2){
if(a[i]==a[i-1] && a[i]==a[i-2]){
fl = true;
}
}
}
if(fl){
cout<<"Yes";
}
else
cout<<"No";
}
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: Given an array A of N integers, find whether there exists three consecutive same integers in the array.The first line of the input contains an integer N.
The second line contains N space separated integers of the array A.
Constraints
3 <= N <= 1000
1 <= A[i] <= 100Output "Yes" if there exists three consecutive equal integers in the array, else output "No" (without quotes).Sample Input
5
1 2 2 2 4
Sample Output
Yes
Explanation: The segment [2, 2, 2] follows the criterion.
Sample Input
5
1 2 2 3 4
Sample Output
No, 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();
}
for(int i=0;i<n-2;i++){
if(a[i]==a[i+1] && a[i+1]==a[i+2]){
System.out.print("Yes");
return;
}
}
System.out.print("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: Given an integer N print the last digit of the given integer.<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>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<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>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: static int LastDigit(int N){
return N%10;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<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>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<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>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: def LastDigit(N):
return N%10, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int x=sc.nextInt();
int ans = 9 * (x-1);
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code:
#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 max1 1000001
#define MOD 1000000000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt[max1];
signed main(){
int n;
cin>>n;
out((n-1)*9);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: a=int(input())
print(9*(a-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long x;
BigInteger sum = new BigInteger("0");
for(int i=0;i<n;i++){
x=sc.nextLong();
sum= sum.add(BigInteger.valueOf(x));
}
sum=sum.divide(BigInteger.valueOf(n));
System.out.print(sum);
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, 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;
signed main() {
IOS;
int n, cur = 0, rem = 0;
cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
cur += (p + rem)/n;
rem = (p + rem)%n;
}
cout << cur;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: n = int(input())
a =list
a=list(map(int,input().split()))
sum=0
for i in range (0,n):
sum=sum+a[i]
print(int(sum//n))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%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 an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, 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,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#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: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, 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);
long n = sc.nextLong();
Long m = sc.nextLong();
if(n%m==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A pair is called lucky if its sum is even and positive. Given three numbers find if there exists a lucky pair or not.The only line contains three integers a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>
<b>Constraints:</b>
-10<sup>9</sup> <= a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> <= 10<sup>9</sup>Print "YES" without quotes if there exists a lucky pair otherwise print "NO" without quotes.Sample Input 1:
23 32 12
Sample Output 1:
YES
Sample Input 2:
1 -1 2
Sample Output 2:
NO, 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>
#pragma GCC target("popcnt")
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 infty = 1e18;
#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 pob pop_back
#define fr first
#define sc 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) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
void solve(){
vector<ll>a(3);
for(ll i = 0;i<3;i++)cin >> a[i];
for(ll i = 0;i<3;i++){
for(ll j = i+1;j<3;j++){
if((a[i]+a[j])>0 && (a[i]+a[j])%2 == 0){
cout << "YES\n";
return;
}
}
}
cout << "NO\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("Input.txt", "r", stdin);
freopen("Output2.txt", "w", stdout);
#endif
ll t = 1;
//cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long count(int[] arr, int l, int h, int[] aux) {
if (l >= h) return 0;
int mid = (l +h) / 2;
long count = 0;
count += count(aux, l, mid, arr);
count += count(aux, mid + 1, h, arr);
count += merge(arr, l, mid, h, aux);
return count;
}
static long merge(int[] arr, int l, int mid, int h, int[] aux) {
long count = 0;
int i = l, j = mid + 1, k = l;
while (i <= mid || j <= h) {
if (i > mid) {
arr[k++] = aux[j++];
} else if (j > h) {
arr[k++] = aux[i++];
} else if (aux[i] <= aux[j]) {
arr[k++] = aux[i++];
} else {
arr[k++] = aux[j++];
count += mid + 1 - i;
}
}
return count;
}
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String str[];
str = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
str = br.readLine().split(" ");
int arr[] =new int[n];
for (int j = 0; j < n; j++) {
arr[j] = Integer.parseInt(str[j]);
}
int[] aux = arr.clone();
System.out.print(count(arr, 0, n - 1, aux));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: count=0
def implementMergeSort(arr,s,e):
global count
if e-s==1:
return
mid=(s+e)//2
implementMergeSort(arr,s,mid)
implementMergeSort(arr,mid,e)
count+=merge_sort_place(arr,s,mid,e)
return count
def merge_sort_place(arr,s,mid,e):
arr3=[]
i=s
j=mid
count=0
while i<mid and j<e:
if arr[i]>arr[j]:
arr3.append(arr[j])
j+=1
count+=(mid-i)
else:
arr3.append(arr[i])
i+=1
while (i<mid):
arr3.append(arr[i])
i+=1
while (j<e):
arr3.append(arr[j])
j+=1
for x in range(len(arr3)):
arr[s+x]=arr3[x]
return count
n=int(input())
arr=list(map(int,input().split()[:n]))
c=implementMergeSort(arr,0,len(arr))
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long _mergeSort(long long arr[], int temp[], int left, int right);
long long merge(long long arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(long long arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(long long arr[], int temp[], int left, int right)
{
long long mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(long long arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;}
int main(){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
long long ans = mergeSort(a, n);
cout << ans; }
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: static void pattern(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;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 a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;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 a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: function pattern(n) {
// write code herenum
for(let i = 1;i<=n;i++){
let str = ''
for(let k = 1; k <= i;k++){
if(k === 1) {
str += `${k}`
}else{
str += ` ${k}`
}
}
console.log(str)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;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 a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: def patternPrinting(n):
for i in range(1,n+1):
for j in range (1,i+1):
print(j,end=' ')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have N coins with either an integer (between 0-9) written on one side and an english letter (a- z) written on the other side.
The following statement must be true for all coins:
<b>If the coin has a vowel on one side, then it must have an even integer on other side. </b>
For example coin having 'b' and '3' is valid (since 'b' is not a vowel, other side can be anything), coin having 'a' and '4' is valid, but coin having 'a' and '5' is invalid.
Now you're given just one side of each coin, find the minimum number of coins you need to flip to check the authenticity of the statement.The first and only line of input contains a string S, where each character in S depicts a side of the coin.
<b>Constraints:</b>
1 ≤ |S| ≤ 50Output a single integer, the minimum number of coins you need to flip.Sample Input
ee
Sample Output
2
Explanation: You need to flip both the coins to make sure an even integer is there on the other side of coin.
Sample Input
0ay1
Sample Output
2, I have written this Solution Code: x = list(input())
c = 0
for i in x:
if i in ['a', 'e', 'i', 'o', 'u', '1', '3', '5', '7', '9']:
c += 1
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have N coins with either an integer (between 0-9) written on one side and an english letter (a- z) written on the other side.
The following statement must be true for all coins:
<b>If the coin has a vowel on one side, then it must have an even integer on other side. </b>
For example coin having 'b' and '3' is valid (since 'b' is not a vowel, other side can be anything), coin having 'a' and '4' is valid, but coin having 'a' and '5' is invalid.
Now you're given just one side of each coin, find the minimum number of coins you need to flip to check the authenticity of the statement.The first and only line of input contains a string S, where each character in S depicts a side of the coin.
<b>Constraints:</b>
1 ≤ |S| ≤ 50Output a single integer, the minimum number of coins you need to flip.Sample Input
ee
Sample Output
2
Explanation: You need to flip both the coins to make sure an even integer is there on the other side of coin.
Sample Input
0ay1
Sample Output
2, 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);
String s = sc.next();
int n = s.length();
int cnt=0;
for(int i=0;i<n;i++){
if(s.charAt(i)>='a' && s.charAt(i)<='z'){
if(s.charAt(i)=='a' || s.charAt(i)=='e' || s.charAt(i)=='o' || s.charAt(i)=='i' ||s.charAt(i)=='u'){
cnt++;
}
}
else{
int x = s.charAt(i)-'0';
if(x%2==1){cnt++;}
}
}
System.out.print(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have N coins with either an integer (between 0-9) written on one side and an english letter (a- z) written on the other side.
The following statement must be true for all coins:
<b>If the coin has a vowel on one side, then it must have an even integer on other side. </b>
For example coin having 'b' and '3' is valid (since 'b' is not a vowel, other side can be anything), coin having 'a' and '4' is valid, but coin having 'a' and '5' is invalid.
Now you're given just one side of each coin, find the minimum number of coins you need to flip to check the authenticity of the statement.The first and only line of input contains a string S, where each character in S depicts a side of the coin.
<b>Constraints:</b>
1 ≤ |S| ≤ 50Output a single integer, the minimum number of coins you need to flip.Sample Input
ee
Sample Output
2
Explanation: You need to flip both the coins to make sure an even integer is there on the other side of coin.
Sample Input
0ay1
Sample Output
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define slld(x) scanf("%lld", &x)
#define pb push_back
#define ll long long
#define mp make_pair
#define F first
#define S second
int main(){
string s;
cin>>s;
int ct=0;
for(int i=0; i<s.length(); i++){
if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u' || s[i]=='1' || s[i]=='3' || s[i]=='5' || s[i]=='7' || s[i]=='9'){
ct++;
}
}
cout<<ct;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N consisting of integers. In one move, you can select any element of the array, and either increase or decrease it by one. In other words, in one move you select an index i (1 ≤ i ≤ N) and replace A<sub>i</sub> by either A<sub>i</sub>-1 or A<sub>i</sub>+1.
For example, if A = [1, 4, -2, 0, 1], if you select i = 3 and decide to increase it, after this move A = [1, 4, -1, 0, 1]. Lets say now you select i = 1 and decrease it, after this move A = [0, 4, -1, 0, 1].
You are also given an integer K in the input. Your task is to find the minimum number of moves you must perform, so that there exist two indices i, j, such that 1 ≤ i ≤ j ≤ N and A[i] + A[i+1] +. . A[j-1] + A[j] = K.The first line contains two integers N and K.
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 ≤ 50000
-10<sup>9</sup> ≤ K ≤ 10<sup>9</sup>
-10<sup>9</sup> ≤ A[i] ≤ 10<sup>9</sup>Print a single integer denoting the minimum number of moves required.Sample Input 1:
3 4
-2 8 2
Sample Output 1:
2
Sample Explanation 1:
Decrement the 2nd element twice, the array become [-2, 6, 2]. Now the segment [1, 2], i. e, [-2, 6] has sum 4., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException
{
StringBuilder out=new StringBuilder();
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] s1=br.readLine().split(" ");
int n=Integer.parseInt(s1[0]);
int k=Integer.parseInt(s1[1]);
String[] s=br.readLine().split(" ");
int[] a=new int[n];
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(s[i]);
}
TreeMap<Long,Long>tm=new TreeMap();
tm.put(0L,1L);
long presum=0,mini=Long.MAX_VALUE;
for(int i=0;i<n;i++)
{
presum+=a[i];
long ch=presum-k;
Long f=tm.floorKey(presum-k);
if(f!=null)
{
if(mini>(ch-f))
{
mini=(ch-f);
}
}
Long s2=tm.ceilingKey(presum-k);
if(s2!=null)
{
if(mini>(s2-ch))
{
mini=(s2-ch);
}
}
tm.put(presum,(long)i);
}
out.append(mini+"\n");
System.out.print(out);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N consisting of integers. In one move, you can select any element of the array, and either increase or decrease it by one. In other words, in one move you select an index i (1 ≤ i ≤ N) and replace A<sub>i</sub> by either A<sub>i</sub>-1 or A<sub>i</sub>+1.
For example, if A = [1, 4, -2, 0, 1], if you select i = 3 and decide to increase it, after this move A = [1, 4, -1, 0, 1]. Lets say now you select i = 1 and decrease it, after this move A = [0, 4, -1, 0, 1].
You are also given an integer K in the input. Your task is to find the minimum number of moves you must perform, so that there exist two indices i, j, such that 1 ≤ i ≤ j ≤ N and A[i] + A[i+1] +. . A[j-1] + A[j] = K.The first line contains two integers N and K.
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 ≤ 50000
-10<sup>9</sup> ≤ K ≤ 10<sup>9</sup>
-10<sup>9</sup> ≤ A[i] ≤ 10<sup>9</sup>Print a single integer denoting the minimum number of moves required.Sample Input 1:
3 4
-2 8 2
Sample Output 1:
2
Sample Explanation 1:
Decrement the 2nd element twice, the array become [-2, 6, 2]. Now the segment [1, 2], i. e, [-2, 6] has sum 4., I have written this Solution Code:
import sys,math,heapq,bisect
input = sys.stdin.readline
sys.setrecursionlimit(10**5)
ints = lambda : list(map(int,input().split()))
p = 10**9+7
inf = 10**20+7
n,k = ints()
a = ints()
ans = inf
s = 0
store = [0]
for i in range(n):
s += a[i]
x = s-k
ind = bisect.bisect_left(store,x)
if ind!=0:
ans = min(ans,abs(x-store[ind-1]))
if ind!=i+1:
ans = min(ans,abs(x-store[ind]))
bisect.insort(store,s)
print(ans), 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 consisting of integers. In one move, you can select any element of the array, and either increase or decrease it by one. In other words, in one move you select an index i (1 ≤ i ≤ N) and replace A<sub>i</sub> by either A<sub>i</sub>-1 or A<sub>i</sub>+1.
For example, if A = [1, 4, -2, 0, 1], if you select i = 3 and decide to increase it, after this move A = [1, 4, -1, 0, 1]. Lets say now you select i = 1 and decrease it, after this move A = [0, 4, -1, 0, 1].
You are also given an integer K in the input. Your task is to find the minimum number of moves you must perform, so that there exist two indices i, j, such that 1 ≤ i ≤ j ≤ N and A[i] + A[i+1] +. . A[j-1] + A[j] = K.The first line contains two integers N and K.
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 ≤ 50000
-10<sup>9</sup> ≤ K ≤ 10<sup>9</sup>
-10<sup>9</sup> ≤ A[i] ≤ 10<sup>9</sup>Print a single integer denoting the minimum number of moves required.Sample Input 1:
3 4
-2 8 2
Sample Output 1:
2
Sample Explanation 1:
Decrement the 2nd element twice, the array become [-2, 6, 2]. Now the segment [1, 2], i. e, [-2, 6] has sum 4., I have written this Solution Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
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>;
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define int long long
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
const int mod = 1e9 + 7;
// const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
vector<pair<int,int>> adj;
void solv()
{
int n, k;
cin>>n>>k;
assert(1<=n && n<=50000);
assert(-1e9<=k && k<=1e9);
readv(a,n);
for(int i = 0;i<n;i++)
assert(-1e9<=a[i] && a[i]<=1e9);
set<int> pref_sm;
pref_sm.insert(0);
int sm = 0;
int mn = INF;
for(int i = 0;i<n;i++)
{
sm += a[i];
int needed = sm - k;
auto p = pref_sm.lower_bound(needed);
if(p!=pref_sm.end())
{
int x = *p;
int diff = x - needed;
assert(diff>=0);
mn = min(mn, diff);
}
if(p!=pref_sm.begin())
{
p--;
int x = *p;
int diff = needed - x;
assert(diff>=0);
mn = min(mn, diff);
}
pref_sm.insert(sm);
}
cout<<mn<<endl;
}
void solve()
{
int t = 1;
// cin>>t;
while(t--)
{
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifdef ONLINE_JUDGE
#ifdef ASC
namespace fs = std::filesystem;
std::string path = "./";
string filename;
for (const auto & entry : fs::directory_iterator(path)){
if( entry.path().extension().string() == ".in"){
filename = entry.path().filename().stem().string();
}
}
if(filename != ""){
string input_file = filename +".in";
string output_file = filename +".out";
if (fopen(input_file.c_str(), "r"))
{
freopen(input_file.c_str(), "r", stdin);
freopen(output_file.c_str(), "w", stdout);
}
}
#endif
#endif
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: static void printInteger(int N){
System.out.println(N);
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printInteger(int x){
printf("%d", x);
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printIntger(int n)
{
cout<<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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: n=int(input())
print (n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<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>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code:
// X and Y are numbers
// ignore number of testcases variable
function pow(X, Y) {
// write code here
// console.log the output in a single line,like example
console.log(Math.pow(X, Y).toFixed(2))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: def power(N,K):
return ("{0:.2f}".format(N**K))
T=int(input())
for i in range(T):
X,N = map(float,input().strip().split())
print(power(X,N)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws Exception {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
double X = Double.parseDouble(str[0]);
int N = Integer.parseInt(str[1]);
System.out.println(String.format("%.2f", myPow(X, N)));
}
}
public static double myPow(double x, int n) {
if (n == Integer.MIN_VALUE)
n = - (Integer.MAX_VALUE - 1);
if (n == 0)
return 1.0;
else if (n < 0)
return 1 / myPow(x, -n);
else if (n % 2 == 1)
return x * myPow(x, n - 1);
else {
double sqrt = myPow(x, n / 2);
return sqrt * sqrt;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ld long double
#define int long long int
#define speed ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
ld power(ld x, ld n){
if(n == 0)
return 1;
else
return x*power(x, n-1);
}
signed main() {
speed;
int t; cin >> t;
while(t--){
double x; int n;
cin >> x >> n;
if(n < 0)
x = 1.0/x, n *= -1;
cout << setprecision(2) << fixed << power(x, n) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code:
int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code: static int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code: int DragonSlayer(int A, int B, int C,int D){
int x = C/B;
if(C%B!=0){x++;}
int y = A/D;
if(A%D!=0){y++;}
if(x<y){return 0;}
return 1;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Natsu is fighting with a dragon that has A Health and B attack power and Natsu has C health and D attack power. The fight goes in turns first Natsu will attack the Dragon then Dragon will attack Natsu and this goes on. The fight will stop when either the dragon's or Natsu's health drops zero or below. Your task is to check whether Natsu will able to slay the Dragon 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>DragonSlayer()</b> that takes integers A, B, C, and D as arguments.
Constraints:-
1 <= A, B, C, D <= 1000Return 0 if Dragon wins else return 1.Sample Input:-
8 2 5 3
Sample Output:-
1
Explanation:-
Natsu's attack:- A = 5, B = 2, C = 5, D = 3
Dragon's attack:- A = 5, B = 2, C = 3, D =3
Natsu's attack:- A = 2, B =2, C = 3, D=3
Dragon's attack:- A = 2, B =2, C = 1, D=3
Natsu's attack:- A = -1, B =2, C = 1, D=3
Natsu's win, I have written this Solution Code:
def DragonSlayer(A,B,C,D):
x = C//B
if(C%B!=0):
x=x+1
y = A//D
if(A%D!=0):
y=y+1
if(x<y):
return 0
return 1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: function Charity(n,m) {
// write code here
// do no console.log the answer
// return the output using return keyword
const per = Math.floor(m / n)
return per > 1 ? per : -1
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: static int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: def Charity(N,M):
x = M//N
if x<=1:
return -1
return x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<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>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Black widow is on a mission to get more information about hydra's whereabouts. She comes across a room where laser lines intersect to form containers. A message pops up on the screen in front of her mocking her saying that she won't be able to find the information she needs. But there's a hint too, the container with maximum water has the intel she needs. The water is polluted with a substance that causes it to become invisible so that no one can make out by seeing how much water is in the container. Help Black Widow solve the mystery and get more intel about hydra.
Find the largest rectangular area that contains most water in a given histogram where the largest rectangle can be made of a number of contiguous bars. For simplicity, assume that all bars have same width and the width is 1 unit.First line contains an integer 'N' denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array. The elements of the array represents the height of the bars.
Constraints:
1 <= N <= 10^5
1 <= A[i] <= 10^9Print a single line denoting the maximum rectangular area possible from the contiguous bars.Input
7
6 2 5 4 5 1 6
Output
12
Input
4
6 3 4 2
Output
9, 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[] st = br.readLine().split("\\s+");
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(st[i]);
}
int max=Integer.MIN_VALUE;
for(int i=0;i<n;i++){
int j=i;
while(j>-1 && arr[i]<=arr[j]){
j--;
}
int k=j+1;
int l=i;
while(l<n && arr[i]<=arr[l]){
l++;
}
int m = l-1;
int total = m-k+1;
int area = total*arr[i];
if(area>max){
max=area;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Black widow is on a mission to get more information about hydra's whereabouts. She comes across a room where laser lines intersect to form containers. A message pops up on the screen in front of her mocking her saying that she won't be able to find the information she needs. But there's a hint too, the container with maximum water has the intel she needs. The water is polluted with a substance that causes it to become invisible so that no one can make out by seeing how much water is in the container. Help Black Widow solve the mystery and get more intel about hydra.
Find the largest rectangular area that contains most water in a given histogram where the largest rectangle can be made of a number of contiguous bars. For simplicity, assume that all bars have same width and the width is 1 unit.First line contains an integer 'N' denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array. The elements of the array represents the height of the bars.
Constraints:
1 <= N <= 10^5
1 <= A[i] <= 10^9Print a single line denoting the maximum rectangular area possible from the contiguous bars.Input
7
6 2 5 4 5 1 6
Output
12
Input
4
6 3 4 2
Output
9, I have written this Solution Code:
def max_area_histogram(histogram):
stack = list()
max_area = 0 # Initialize max area
index = 0
while index < len(histogram):
if (not stack) or (histogram[stack[-1]] <= histogram[index]):
stack.append(index)
index += 1
else:
top_of_stack = stack.pop()
area = (histogram[top_of_stack] *
((index - stack[-1] - 1)
if stack else index))
max_area = max(max_area, area)
while stack:
top_of_stack = stack.pop()
area = (histogram[top_of_stack] *
((index - stack[-1] - 1)
if stack else index))
max_area = max(max_area, area)
return max_area
n = int(input())
arr = list(map(int, input().split()))
print(max_area_histogram(arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Black widow is on a mission to get more information about hydra's whereabouts. She comes across a room where laser lines intersect to form containers. A message pops up on the screen in front of her mocking her saying that she won't be able to find the information she needs. But there's a hint too, the container with maximum water has the intel she needs. The water is polluted with a substance that causes it to become invisible so that no one can make out by seeing how much water is in the container. Help Black Widow solve the mystery and get more intel about hydra.
Find the largest rectangular area that contains most water in a given histogram where the largest rectangle can be made of a number of contiguous bars. For simplicity, assume that all bars have same width and the width is 1 unit.First line contains an integer 'N' denoting the size of array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array. The elements of the array represents the height of the bars.
Constraints:
1 <= N <= 10^5
1 <= A[i] <= 10^9Print a single line denoting the maximum rectangular area possible from the contiguous bars.Input
7
6 2 5 4 5 1 6
Output
12
Input
4
6 3 4 2
Output
9, 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 l[N], r[N], a[N];
signed main() {
IOS;
int n; cin >> n;
stack<int> st;
st.push(0);
for(int i = 1; i <= n; i++){
cin >> a[i];
while(!st.empty() && a[st.top()] >= a[i])
st.pop();
l[i] = st.top()+1;
st.push(i);
}
while(!st.empty())
st.pop();
st.push(n+1);
for(int i = n; i >= 1; i--){
while(!st.empty() && a[st.top()] >= a[i])
st.pop();
r[i] = st.top()-1;
st.push(i);
}
int ans = 0;
for(int i = 1; i <= n; i++)
ans = max(ans, (r[i]-l[i]+1)*a[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 A, find the nearest smaller element S[i] for every element A[i] in the array such that the element has an index smaller than i.
More formally,
S[i] for an element A[i] = an element A[j] such that
j is maximum possible AND
j < i AND
A[j] <= A[i]
Elements for which no smaller element exist, consider next smaller element as -1.The first line contains the size of array, n
The next line n elements of the integer array, A[i]
<b>Constraints:</b>
1 <= n <= 10^5
1 <= A[i] <= 10^6Print the integer array S such that S[i] contains nearest smaller number than A[i] such than index of S[i] is less than 'i'. If no such element occurs S[i] should be -1.Input:
5
4 5 2 10 8
Output:
-1 4 -1 2 2
Explanation 1:
index 1: No element less than 4 in left of 4, G[1] = -1
index 2: A[1] is only element less than A[2], G[2] = A[1]
index 3: No element less than 2 in left of 2, G[3] = -1
index 4: A[3] is nearest element which is less than A[4], G[4] = A[3]
index 4: A[3] is nearest element which is less than A[5], G[5] = A[3]
Input:
5
1 2 3 4 5
Output:
-1 1 2 3 4, I have written this Solution Code: size = int(input())
arr = list(map(int,input().split()))
tempArr = arr.copy()
def nearLeast(pos):
for i in range(pos-1,-1,-1):
if(tempArr[i]<=tempArr[pos]):
return tempArr[i]
return -1
for i in range(size):
if(i==0):
arr[0] = -1
else:
arr[i] = nearLeast(i)
print(*arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A, find the nearest smaller element S[i] for every element A[i] in the array such that the element has an index smaller than i.
More formally,
S[i] for an element A[i] = an element A[j] such that
j is maximum possible AND
j < i AND
A[j] <= A[i]
Elements for which no smaller element exist, consider next smaller element as -1.The first line contains the size of array, n
The next line n elements of the integer array, A[i]
<b>Constraints:</b>
1 <= n <= 10^5
1 <= A[i] <= 10^6Print the integer array S such that S[i] contains nearest smaller number than A[i] such than index of S[i] is less than 'i'. If no such element occurs S[i] should be -1.Input:
5
4 5 2 10 8
Output:
-1 4 -1 2 2
Explanation 1:
index 1: No element less than 4 in left of 4, G[1] = -1
index 2: A[1] is only element less than A[2], G[2] = A[1]
index 3: No element less than 2 in left of 2, G[3] = -1
index 4: A[3] is nearest element which is less than A[4], G[4] = A[3]
index 4: A[3] is nearest element which is less than A[5], G[5] = A[3]
Input:
5
1 2 3 4 5
Output:
-1 1 2 3 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;
int n; cin >> n;
stack<int> s;
for(int i = 1; i <= n; i++){
cin >> a[i];
while(!s.empty() && a[s.top()] > a[i])
s.pop();
if(s.empty())
cout << -1 << " ";
else
cout << a[s.top()] << " ";
s.push(i);
}
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, find the nearest smaller element S[i] for every element A[i] in the array such that the element has an index smaller than i.
More formally,
S[i] for an element A[i] = an element A[j] such that
j is maximum possible AND
j < i AND
A[j] <= A[i]
Elements for which no smaller element exist, consider next smaller element as -1.The first line contains the size of array, n
The next line n elements of the integer array, A[i]
<b>Constraints:</b>
1 <= n <= 10^5
1 <= A[i] <= 10^6Print the integer array S such that S[i] contains nearest smaller number than A[i] such than index of S[i] is less than 'i'. If no such element occurs S[i] should be -1.Input:
5
4 5 2 10 8
Output:
-1 4 -1 2 2
Explanation 1:
index 1: No element less than 4 in left of 4, G[1] = -1
index 2: A[1] is only element less than A[2], G[2] = A[1]
index 3: No element less than 2 in left of 2, G[3] = -1
index 4: A[3] is nearest element which is less than A[4], G[4] = A[3]
index 4: A[3] is nearest element which is less than A[5], G[5] = A[3]
Input:
5
1 2 3 4 5
Output:
-1 1 2 3 4, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int arrSize = sc.nextInt();
int arr[] = new int[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextInt();
nearSmaller(arr, arrSize);
}
static void nearSmaller(int arr[], int arrSize)
{
Stack<Integer> s = new Stack<>();
for(int i = 0; i < arrSize; i++)
{
while(!s.empty() == true && arr[s.peek()] > arr[i])
s.pop();
if(s.empty() == true)
System.out.print(-1 +" ");
else
System.out.print(arr[s.peek()]+" ");
s.push(i);
}
}
}, 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 digits between 1 and 9. Rearrange the string such that the number of even substrings it contains is maximised. A substring is said to be even if the number it represents is even.<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>solve()</b> that takes the string S as parameter.
<b>Constraints</b>
1<=|S|<=1000
S only contains digits between 1 and 9 (both inclusive).Return the number of even substrings present in the string after rearranging it such that the number is maximised.Sample Input:
1232
Sample Output:
7
We will rearrange the string to "1322". Now, the seven even substrings are "132", "32", "2", "2", "1232", "232", "32". Two substrings are considered different if at least one of the starting or ending index differs from the other., I have written this Solution Code:
class Solution {
public long solve(String S) {
int e = 0;
for (int i = 0; i < S.length(); i++) {
int y = S.charAt(i) - '0';
if (y % 2 == 0) {
e++;
}
}
long ans = 0;
for (int i = S.length(); i >= 0; i--) {
if (e > 0) {
ans += i;
e--;
}
}
return ans;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code:
// X and Y are numbers
// ignore number of testcases variable
function pow(X, Y) {
// write code here
// console.log the output in a single line,like example
console.log(Math.pow(X, Y).toFixed(2))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: def power(N,K):
return ("{0:.2f}".format(N**K))
T=int(input())
for i in range(T):
X,N = map(float,input().strip().split())
print(power(X,N)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws Exception {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
double X = Double.parseDouble(str[0]);
int N = Integer.parseInt(str[1]);
System.out.println(String.format("%.2f", myPow(X, N)));
}
}
public static double myPow(double x, int n) {
if (n == Integer.MIN_VALUE)
n = - (Integer.MAX_VALUE - 1);
if (n == 0)
return 1.0;
else if (n < 0)
return 1 / myPow(x, -n);
else if (n % 2 == 1)
return x * myPow(x, n - 1);
else {
double sqrt = myPow(x, n / 2);
return sqrt * sqrt;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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