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For this Question: Given an integer array where every element appears an even number of times, except one element which appears an odd number of times. If the identical elements appear in pairs in the array and there cannot be more than two consecutive occurrences of an element, find the odd occurring element.The first line contains a single integer N.
The second line contains N space- separated integer A[i].
<b>Constraints</b>
1 ≤ N ≤ 10<sup>4</sup>
1 ≤ arr[i] ≤ 10<sup>3</sup>Print odd occurring element.Sample Input 1:
5
5 5 3 4 4
Sample Output 1:
3
Sample Input 2:
5
1 1 4 4 16
Sample Output 2:
16, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
d={}
for i in arr:
d[i]=d.get(i,0)+1
print([k for k,v in d.items() if v==min(d.values())][0]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array where every element appears an even number of times, except one element which appears an odd number of times. If the identical elements appear in pairs in the array and there cannot be more than two consecutive occurrences of an element, find the odd occurring element.The first line contains a single integer N.
The second line contains N space- separated integer A[i].
<b>Constraints</b>
1 ≤ N ≤ 10<sup>4</sup>
1 ≤ arr[i] ≤ 10<sup>3</sup>Print odd occurring element.Sample Input 1:
5
5 5 3 4 4
Sample Output 1:
3
Sample Input 2:
5
1 1 4 4 16
Sample Output 2:
16, I have written this Solution Code: #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int findOddOccuring(vector<int> nums)
{
sort(nums.begin(), nums.end());
int i = 0;
while (i < nums.size())
{
int curr = nums[i];
int count = 0;
while (i < nums.size() && nums[i] == curr)
{
count++;
i++;
}
if (count & 1) {
return curr;
}
}
return -1;
}
int main()
{
int n;
cin>>n;
vector<int>nums(n);
for(int &i:nums)cin>>i;
cout <<findOddOccuring(nums);
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: 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: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code:
import sys
from collections import defaultdict
from heapq import heappush, heappop
n, m = map(int, input().split())
d = defaultdict(list)
dist = [sys.maxsize]*n
dist[0] = 0
for _ in range(m):
start, dest, wt = map(int, input().split())
d[start-1].append((dest-1, wt))
d[dest-1].append((start-1, wt))
heap = [(0, 0, 0)]
while heap:
count, cost, u= heappop(heap)
for vertex, weight in d[u]:
if dist[vertex] > cost + weight*(count+1):
dist[vertex] = cost + weight*(count+1)
heappush(heap, (count+1, dist[vertex], vertex))
for d in dist:
if d == sys.maxsize:
print(-1)
else:
print(d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF =1e18;
vector<tuple<int, int, int>> adj;
void solve()
{
int n, m;
cin>>n>>m;
// assert(1<=n && n<=3000);
// assert(0<=m && m<=10000);
adj.resize(n);
for(int i = 0;i<m;i++)
{
int x, y, w;
cin>>x>>y>>w;
x--;
y--;
// assert(0<=x && x<n);
// assert(0<=y && y<n);
// assert(1<=w && w<=1e9);
adj.push_back({x, y, w});
adj.push_back({y, x, w});
}
vector<int> dp_old(n, INF);
vector<int> dp_new(n, INF);
dp_old[0] = 0;
for(int i = 1;i<=n;i++)
{
fill(dp_new.begin(), dp_new.end(), INF);
for(auto [x, y,w]:adj)
{
dp_new[y]= min({dp_new[y], dp_old[x] + i * w});
}
for(int j = 0;j<n;j++)
dp_new[j] = min(dp_new[j], dp_old[j]);
swap(dp_new, dp_old);
}
for(int i = 0;i<n;i++)
{
if(dp_old[i] == INF)
dp_old[i] = -1;
cout<<dp_old[i]<<"\n";
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq()throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=1;
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
int m=i();
LinkedList<int[]> l[]=new LinkedList[n];
for(int x=0;x<n;x++)l[x]=new LinkedList<>();
for(int x=0;x<m;x++)
{
int a=i()-1;
int b=i()-1;
int c=i();
l[a].add(new int[]{b,c});
l[b].add(new int[]{a,c});
}
long d[]=new long[n];
for(int x=0;x<n;x++)d[x]=maxl;
d[0]=0l;
PriorityQueue<long[]> p=new PriorityQueue<>(5000,(a,b)->a[2]-b[2]<1l?-1:1);
p.add(new long[]{0l,0,0});
while(p.size()>0)
{
long r[]=p.poll();
long di=r[0];
int no=(int)r[1];
long mu=r[2];
for(int e[]:l[no])
{
int node=e[0];
int w=e[1];
long de=di+w*(mu+1);
if(d[node]>de)
{
d[node]=de;
p.add(new long[]{de,node,mu+1});
}
}
}
for(long x:d)
{
sl(x==maxl?-1:x);
}
}
p(sb);
}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
StringBuilder sb;public static void main(String[] a)throws Exception{new Main().tq();}int[] so(int ar[])
{Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;
x++)ar[x]=r[x];return ar;}long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)
r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[]) {Character
r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)
ar[x]=r[x];return ar;}void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.
append(s);}void s(char s){sb.append(s);}void s(double s){sb.append(s);}void ss(){sb.append(' ');}void sl
(String s){sb.append(s);sb.append("\n");}void sl(int s){sb.append(s);sb.append("\n");}void sl(long s){sb
.append(s);sb.append("\n");}void sl(char s) {sb.append(s);sb.append("\n");}void sl(double s){sb.append(s)
;sb.append("\n");}void sl(){sb.append("\n");}int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
{return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
{while(b>0){int c=a%b;a=b;b=c;}return a;}boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=
s.charAt(j--))return false;return true;}boolean[] si(int n) {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]
=true;for(int x=4;x<=n;x+=2)bo[x]=true;for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;
y+=vv)bo[y]=true;}}return bo;}long mul(long a,long b,long m) {long r=1l;a%=m;while(b>0){if((b&1)==1)
r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}int i()throws IOException{if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
{out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
{out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
println(p);}void pl(boolean p)
{out.println(p);}void pl(){out.println();}void s(int a[]){for(int e:a)
{sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[])
{for(long e:a){sb.append(e);sb.append(' ')
;}sb.append("\n");}void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append
("\n");}}
void s(char a[])
{for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}void s(char ar[][])
{for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}int[] ari(int n)throws
IOException {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());for(int x=0;
x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}int[][] ari(int n,int m)throws
IOException {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if (!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}long[] arl
(int n)throws IOException {long ar[]=new long[n];if(!st.hasMoreTokens()) st=new StringTokenizer(bq.readLine())
;for(int x=0;x<n;x++)ar[x]=Long.parseLong(st.nextToken());return ar;}long[][] arl(int n,int m)throws
IOException {long ar[][]=new long[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens()) st=new
StringTokenizer(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException {String ar[] =new String[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken();return ar;}double[] ard
(int n)throws IOException {double ar[] =new double[n];if(!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}double[][] ard
(int n,int m)throws IOException{double ar[][]=new double[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens())
st=new StringTokenizer(bq.readLine());for(int y=0;y<m;y++) ar[x][y]=Double.parseDouble(st.nextToken());}
return ar;}char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}char[][]
arc(int n,int m)throws IOException {char ar[][]=new char[n][m];for(int x=0;x<n;x++){String s=bq.readLine();
for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}void p(int ar[])
{StringBuilder sb=new StringBuilder
(2*ar.length);for(int a:ar){sb.append(a);sb.append(' ');}out.println(sb);}void p(int ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(int a[]:ar){for(int aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(long ar[]){StringBuilder sb=new StringBuilder
(2*ar.length);for(long a:ar){ sb.append(a);sb.append(' ');}out.println(sb);}
void p(long ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(long a[]:ar){for(long aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;
StringBuilder sb=new StringBuilder(c);for(String a:ar){sb.append(a);sb.append(' ');}out.println(sb);}
void p(double ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
sb.append(' ');}out.println(sb);}void p
(double ar[][]){StringBuilder sb=new StringBuilder(2*
ar.length*ar[0].length);for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n")
;}p(sb);}void p(char ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(char aa:ar){sb.append(aa);
sb.append(' ');}out.println(sb);}void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0]
.length);for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, 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;
int ans = 0;
For(i, 0, n){
int a; cin>>a;
ans += a;
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code: n = int(input())
chocolates = list(map(int, input().strip().split(" ")))
count = 0
for val in chocolates:
count += val
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
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);
int n = sc.nextInt();
int A[] = new int[n];
for(int i=0;i<n;i++){
A[i]=sc.nextInt();
}
int Total=0;
for(int i=0;i<n;i++){
Total+=A[i];
}
System.out.print(Total);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write an algorithm to determine if a number n is happy. A happy number is a number defined by the following process:<ul><li>Starting with any positive integer, replace the number by the sum of the squares of its digits. </li><li>Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. </li><li>Those numbers for which this process ends in 1 are happy. </li></ul>Return true if n is a happy number, and false if not.The first line of the input contains the number n.
Constraints
1 <= n <= 2<sup>31</sup>-1Print <b>true</b> if it's a happy number otherwise <b>false</b>.Sample Input
19
Sample Output
true
Explanation
1<sup>2</sup> + 9<sup>2</sup> = 82
8 <sup>2</sup> + 2<sup>2</sup> = 68
6<sup>2</sup> + 8<sup>2</sup> = 100
1<sup>2</sup> + 0<sup>2</sup> + 0<sup>2</sup> = 1, I have written this Solution Code: from collections import defaultdict
def sum_int(num):
tot=0
while(num!=0):
tot+=(num%10)**2
num=num//10
return tot
n=int(input())
d=defaultdict(int)
while(n!=1 and d[n]==0):
d[n]+=1
n=sum_int(n)
if(n==1):
print("true")
else:
print("false")
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write an algorithm to determine if a number n is happy. A happy number is a number defined by the following process:<ul><li>Starting with any positive integer, replace the number by the sum of the squares of its digits. </li><li>Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. </li><li>Those numbers for which this process ends in 1 are happy. </li></ul>Return true if n is a happy number, and false if not.The first line of the input contains the number n.
Constraints
1 <= n <= 2<sup>31</sup>-1Print <b>true</b> if it's a happy number otherwise <b>false</b>.Sample Input
19
Sample Output
true
Explanation
1<sup>2</sup> + 9<sup>2</sup> = 82
8 <sup>2</sup> + 2<sup>2</sup> = 68
6<sup>2</sup> + 8<sup>2</sup> = 100
1<sup>2</sup> + 0<sup>2</sup> + 0<sup>2</sup> = 1, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
private static int getNext(int n) {
int totalSum = 0;
while (n > 0) {
int d = n % 10;
n = n / 10;
totalSum += d * d;
}
return totalSum;
}
public static boolean isHappy(int n) {
Set<Integer> seen = new HashSet<>();
while (n != 1 && !seen.contains(n)) {
seen.add(n);
n = getNext(n);
}
return n == 1;
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner inp = new Scanner(System.in);
int n = inp.nextInt();
System.out.println(isHappy(n));
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<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>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: def MaximumProduct(N):
ans=N//4
if N %4 !=0:
ans=ans+1
return ans, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<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>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<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>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<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>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: static int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<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 integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code: def SortPair(items,n):
items.sort(reverse = True)
return items, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code:
static Pair[] SortPair(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
if(p1.x==p2.x){
return p2.y-p1.y;
}
return p2.x-p1.x;
}
});
return arr;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, 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: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two integer arrays A and B of sizes N and M respectively. You need to modify the elements of A so that B becomes its subarray. Modifying an element means change the element to any other value.
Find the minimum number of modifications to achieve this.The first line of the input contains two integers N and M.
The second line of the input contains N space separated integers, the elements of array A.
The third line of the input contains M space separated integers, the elements of array B.
Constraints
1 <= M <= N <= 500
1 <= A[i], B[i] <= 10Output a single integer, the minimum number of modifications in A to make B its subarray.Sample Input
6 3
3 1 2 1 3 3
1 2 3
Sample Output
1
Explanation: If you modify A[4] from 1 to 3. A[2]. A[4] represents the array B, so B is its subarray.
Sample Input
10 5
3 4 5 3 4 3 1 3 5 2
1 4 4 4 3
Sample Output
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int m = Integer.parseInt(str[1]);
int listn[] = new int[n];
int listm[] = new int[m];
str = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
listn[i]=Integer.parseInt(str[i]);
str = read.readLine().trim().split(" ");
for(int i = 0; i < m; i++)
listm[i]=Integer.parseInt(str[i]);
int ans =m;
for(int i=0; i < n-m+1 ; i++){
int ct=0;
for(int j=0;j <m; j++){
if(listn[i+j] != listm[j]){
ct++;
}
}
ans = Math.min(ans,ct);
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two integer arrays A and B of sizes N and M respectively. You need to modify the elements of A so that B becomes its subarray. Modifying an element means change the element to any other value.
Find the minimum number of modifications to achieve this.The first line of the input contains two integers N and M.
The second line of the input contains N space separated integers, the elements of array A.
The third line of the input contains M space separated integers, the elements of array B.
Constraints
1 <= M <= N <= 500
1 <= A[i], B[i] <= 10Output a single integer, the minimum number of modifications in A to make B its subarray.Sample Input
6 3
3 1 2 1 3 3
1 2 3
Sample Output
1
Explanation: If you modify A[4] from 1 to 3. A[2]. A[4] represents the array B, so B is its subarray.
Sample Input
10 5
3 4 5 3 4 3 1 3 5 2
1 4 4 4 3
Sample Output
3, I have written this Solution Code: n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
k = 0
matched = 0
while(n-k >= m):
c = 0
for i in range(m):
if(a[k:][i] == b[i]):
c += 1
matched = max(matched,c)
k += 1
print(m-matched), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two integer arrays A and B of sizes N and M respectively. You need to modify the elements of A so that B becomes its subarray. Modifying an element means change the element to any other value.
Find the minimum number of modifications to achieve this.The first line of the input contains two integers N and M.
The second line of the input contains N space separated integers, the elements of array A.
The third line of the input contains M space separated integers, the elements of array B.
Constraints
1 <= M <= N <= 500
1 <= A[i], B[i] <= 10Output a single integer, the minimum number of modifications in A to make B its subarray.Sample Input
6 3
3 1 2 1 3 3
1 2 3
Sample Output
1
Explanation: If you modify A[4] from 1 to 3. A[2]. A[4] represents the array B, so B is its subarray.
Sample Input
10 5
3 4 5 3 4 3 1 3 5 2
1 4 4 4 3
Sample Output
3, 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, m;
cin>>n>>m;
int a[n], b[m];
For(i, 0, n){
cin>>a[i];
}
For(i, 0, m){
cin>>b[i];
}
int ans = m;
For(i, 0, n-m+1){
int ct = 0;
For(j, 0, m){
if(a[i+j]!=b[j])
ct++;
}
ans = min(ans, ct);
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: def partition(array, low, high):
pivot = array[high]
i = low - 1
for j in range(low, high):
if array[j] <= pivot:
i = i + 1
(array[i], array[j]) = (array[j], array[i])
(array[i + 1], array[high]) = (array[high], array[i + 1])
return i + 1
def quick_sort(array, low, high):
if low < high:
pi = partition(array, low, high)
quick_sort(array, low, pi - 1)
quick_sort(array, pi + 1, high)
t=int(input())
for i in range(t):
n=int(input())
a=input().strip().split()
a=[int(i) for i in a]
quick_sort(a, 0, n - 1)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] quickSort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
public static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: function quickSort(arr, low, high)
{
if(low < high)
{
let pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
function partition(arr, low, high)
{
let pivot = arr[high];
let i = (low-1); // index of smaller element
for (let j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
let temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}, In this Programming Language: JavaScript, 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: You are given an array <b>A</b> of size <b>N</b>. You need to find the integer not present in this array that is closest to <b>K</b>. The closeness between two numbers A[i] and A[j] is defined as <b>abs(A[i] - A[j])</b>.
For Example:
For given array A = [1,2,5,6,13]
For given K = 12
Output: 12
Explanation: numbers missing 3, 4, 7,8, 9, 10, 11, 12, 14, 15....
and K = 12, so distance from each missing number is (3,9), (4,8), ....
where every pair denotes (missing number, distance) hence 12 is answer.The first line of the input contains two integers N and K.
Constraints
1 <= N <= 1000
1 <= A[i] <= 10000
1 <= K <= 10000
(The array may not contain all distinct numbers)Output the required closest number.Sample Input
5 6
4 7 10 6 5
Sample Output
8
Explanation: Closeness of 8 is 2. There is no such number that has closeness lesser than this and is not a part of array.
Sample Input
5 10
4 7 10 6 5
Sample 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));
String[] str=br.readLine().split(" ");
int N=Integer.parseInt(str[0]);
int K=Integer.parseInt(str[1]);
str=br.readLine().split(" ");
int[] arr=new int[N];
for(int i=0;i<N;i++)
{
arr[i]=Integer.parseInt(str[i]);
}
Arrays.sort(arr);
int i=0;
boolean flag=false;
int index=binarySearch(arr,0,N-1,K-i);
while(index!=-1)
{
i++;
index=binarySearch(arr,0,index,K-i);
if(index!=-1)
{
index=binarySearch(arr,index,N-1,K+i);
}
else
flag=true;
}
if(flag)
System.out.print(K-i);
else
System.out.print(K+i);
}
public static int binarySearch(int[] arr,int start,int end,int key)
{
while(start<=end)
{
int mid=(start+end)/2;
if(arr[mid]==key)
return mid;
else if(arr[mid]>key)
end=mid-1;
else
start=mid+1;
}
return -1;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array <b>A</b> of size <b>N</b>. You need to find the integer not present in this array that is closest to <b>K</b>. The closeness between two numbers A[i] and A[j] is defined as <b>abs(A[i] - A[j])</b>.
For Example:
For given array A = [1,2,5,6,13]
For given K = 12
Output: 12
Explanation: numbers missing 3, 4, 7,8, 9, 10, 11, 12, 14, 15....
and K = 12, so distance from each missing number is (3,9), (4,8), ....
where every pair denotes (missing number, distance) hence 12 is answer.The first line of the input contains two integers N and K.
Constraints
1 <= N <= 1000
1 <= A[i] <= 10000
1 <= K <= 10000
(The array may not contain all distinct numbers)Output the required closest number.Sample Input
5 6
4 7 10 6 5
Sample Output
8
Explanation: Closeness of 8 is 2. There is no such number that has closeness lesser than this and is not a part of array.
Sample Input
5 10
4 7 10 6 5
Sample Output
9, I have written this Solution Code: size,k = map(int,input().split())
lis = list(map(int,input().split()))
l=[]
m=0
for i in lis:
if(k+m in lis and k-m in lis):
m=m+1
else:
if(k-m in lis):
print(k+m)
else:
print(k-m)
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array <b>A</b> of size <b>N</b>. You need to find the integer not present in this array that is closest to <b>K</b>. The closeness between two numbers A[i] and A[j] is defined as <b>abs(A[i] - A[j])</b>.
For Example:
For given array A = [1,2,5,6,13]
For given K = 12
Output: 12
Explanation: numbers missing 3, 4, 7,8, 9, 10, 11, 12, 14, 15....
and K = 12, so distance from each missing number is (3,9), (4,8), ....
where every pair denotes (missing number, distance) hence 12 is answer.The first line of the input contains two integers N and K.
Constraints
1 <= N <= 1000
1 <= A[i] <= 10000
1 <= K <= 10000
(The array may not contain all distinct numbers)Output the required closest number.Sample Input
5 6
4 7 10 6 5
Sample Output
8
Explanation: Closeness of 8 is 2. There is no such number that has closeness lesser than this and is not a part of array.
Sample Input
5 10
4 7 10 6 5
Sample Output
9, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int x, n; cin>>n>>x;
set<int> s;
For(i, 0, n){
int a; cin>>a;
s.insert(a);
}
int v = INF;
int cur = -INF;
For(i, -1, 20001){
if(s.find(i)!=s.end())
continue;
if(abs(i-x)<v){
cur = i;
v = abs(i-x);
}
}
cout<<cur;
}
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 a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",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 "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
PrintWriter out= new PrintWriter(System.out);
int t=Integer.parseInt(br.readLine());
while(t>0){
t-=1;
int n=Integer.parseInt(br.readLine());
String s[]=br.readLine().split(" ");
int arr[]=new int[n];
int zero_counter=0,one_counter=0,two_counter=0;
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(s[i]);
if(arr[i]==0)
zero_counter+=1;
else if(arr[i]==1)
one_counter+=1;
else
two_counter+=1;
}
for(int i=0;i<zero_counter;i++){
out.print(0+" ");
}
for(int i=0;i<one_counter;i++){
out.print(1+" ");
}
for(int i=0;i<two_counter;i++){
out.print(2+" ");
}
out.flush();
out.println(" ");
}
out.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=input()
a=map(int,input().split())
z=o=tc=0
for i in a:
if i==0:z+=1
elif i==1:o+=1
else:tc+=1
while z>0:
z-=1
print(0,end=' ')
while o>0:
o-=1
print(1,end=' ')
while tc>0:
tc-=1
print(2,end=' ')
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., 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 t; cin >> t;
while(t--){
int n; cin >> n;
int a[3] = {0};
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
a[1] += a[0];
a[2] += a[1];
for(int i = 1; i <= n; i++){
if(i <= a[0]) cout << "0 ";
else if(i <= a[1]) cout << "1 ";
else cout << "2 ";
}
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., I have written this Solution Code: function zeroOneTwoSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
out.println(sumTotient(ni()/ni()));
}
public static int[] enumTotientByLpf(int n, int[] lpf)
{
int[] ret = new int[n+1];
ret[1] = 1;
for(int i = 2;i <= n;i++){
int j = i/lpf[i];
if(lpf[j] != lpf[i]){
ret[i] = ret[j] * (lpf[i]-1);
}else{
ret[i] = ret[j] * lpf[i];
}
}
return ret;
}
public static int[] enumLowestPrimeFactors(int n)
{
int tot = 0;
int[] lpf = new int[n+1];
int u = n+32;
double lu = Math.log(u);
int[] primes = new int[(int)(u/lu+u/lu/lu*1.5)];
for(int i = 2;i <= n;i++)lpf[i] = i;
for(int p = 2;p <= n;p++){
if(lpf[p] == p)primes[tot++] = p;
int tmp;
for(int i = 0;i < tot && primes[i] <= lpf[p] && (tmp = primes[i]*p) <= n;i++){
lpf[tmp] = primes[i];
}
}
return lpf;
}
public static long sumTotient(int n)
{
if(n == 0)return 0L;
if(n == 1)return 1L;
int s = (int)Math.sqrt(n);
long[] cacheu = new long[n/s];
long[] cachel = new long[s+1];
int X = (int)Math.pow(n, 0.66);
int[] lpf = enumLowestPrimeFactors(X);
int[] tot = enumTotientByLpf(X, lpf);
long sum = 0;
int p = cacheu.length-1;
for(int i = 1;i <= X;i++){
sum += tot[i];
if(i <= s){
cachel[i] = sum;
}else if(p > 0 && i == n/p){
cacheu[p] = sum;
p--;
}
}
for(int i = p;i >= 1;i--){
int x = n/i;
long all = (long)x*(x+1)/2;
int ls = (int)Math.sqrt(x);
for(int j = 2;x/j > ls;j++){
long lval = i*j < cacheu.length ? cacheu[i*j] : cachel[x/j];
all -= lval;
}
for(int v = ls;v >= 1;v--){
long w = x/v-x/(v+1);
all -= cachel[v]*w;
}
cacheu[(int)i] = all;
}
return cacheu[1];
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new Main().run(); }
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
private int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Phoebe is a big GCD fan. Being bored, she starts counting number of pairs of integers (A, B) such that following conditions are satisfied:
<li> GCD(A, B) = X (As X is Phoebe's favourite integer)
<li> A <= B <= L
As Phoebe's performance is coming up, she needs your help to find the number of such pairs possible.
Note: GCD refers to the <a href="https://en.wikipedia.org/wiki/Greatest_common_divisor">Greatest common divisor</a>.Input contains two integers L and X.
Constraints:
1 <= L, X <= 1000000000Print a single integer denoting number of pairs possible.Sample Input
5 2
Sample Output
2
Explanation: Pairs satisfying all conditions are: (2, 2), (2, 4)
Sample Input
5 3
Sample Output
1
Explanation: Pairs satisfying all conditions are: (3, 3), I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
ull X[20000001];
ull cmp(ull N){
return N*(N+1)/2;
}
ull solve(ull N){
if(N==1)
return 1;
if(N < 20000001 && X[N] != 0)
return X[N];
ull res = 0;
ull q = floor(sqrt(N));
for(int k=2;k<N/q+1;++k){
res += solve(N/k);
}
for(int m=1;m<q;++m){
res += (N/m - N/(m+1)) * solve(m);
}
res = cmp(N) - res;
if(N < 20000001)
X[N] = res;
return res;
}
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int l,x;
cin>>l>>x;
if(l<x)
cout<<0;
else
cout<<solve(l/x);
#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, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
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, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, 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, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, 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, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 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, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".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 side = scanner.nextInt();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), 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: 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 number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",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 "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <i> No more Rick and Morty :D</i>
Tono and Solo, the small girls, are ready for the great <i> Super Small League</i>.
There are N participants for the league standing in a queue. Each player has an associated happiness H<sub>i</sub> and associated strength S<sub>i</sub>. Now, here's the selection procedure.
Both Tono and Solo take alternate turns. In a turn, the small girl can choose to either select the first player in the queue, doing this increases her happiness by H<sub>i</sub> and her team's strength by S<sub>i</sub>, or not select the first player in the queue, doing this makes the her sad and decreases her happiness by 1 and has no impact on the team's strength.
Note: If a player is chosen, she is removed from the queue (Of course :P) and the selection process stops when the queue becomes empty.
The initial happiness of Tono and Solo are X and Y respectively.
Now, both Tono and Solo want to select their team so as to maximize their team's strength. Can you find the strength of their teams if their happiness should never fall below 0?
As Tono is smaller, she goes first!The first line of the input contains three integers N, X, and Y denoting the number of participants, and the initial happiness of Tono and Solo respectively.
The next lines contains two integers H<sub>i</sub> and S<sub>i</sub>, the happiness received by taking the i-th player in the team, and the strength of the i-th player.
Constraints
1 <= N <= 150
0 <= X, Y, H<sub>i</sub> <= 10<sup>9</sup>
S<sub>i</sub> >= 0
Sum of S<sub>i</sub> of all players <= 150
Output two integers, the strength of Tono's and Solo's team.Sample Input 1
2 5 4
5 6
4 9
Sample Output 1
9 6
Sample Input 2
3 50 1
49 1
0 9
0 1
Sample Output 2
9 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int dp[304][305];
int s[305],sum[304];ll r[305];
void solve(){
int n; ll a,b;
cin>>n>>a>>b;
int tot = 0;
for(int i=1;i<=n;i++) {
cin>>r[i]>>s[i];
tot += s[i];
}
assert(tot <= 150);
sum[n+1]=0;
for(int i=n; i>=1; i--)sum[i]=sum[i+1]+s[i];
for(int j=s[n]+1; j<=300; j++)dp[n][j]=INF;
for(int j=0; j<=s[n]; j++)dp[n][j]=-INF;
for(int i=n-1; i>=1; i--)
{
for(int j=300; j>=0; j--)
{
if(j>sum[i]){
dp[i][j]=INF;
continue;
}
dp[i][j]=dp[i][j+1];
int u=sum[i]-j;
dp[i][j]=min(dp[i][j],-r[i]-dp[i+1][u+1]+1);
if(j <= sum[i+1])
{
dp[i][j]=min(dp[i][j], max(1LL, dp[i+1][j]+r[i]+1LL));
}
}
}
for(int j=sum[1];j>=0;j--)if(dp[1][j]<=a-b){cout<<j<<" "<<sum[1]-j<<endl;return;}
}
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 two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 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: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: def profit(C, S):
print(S - C), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: static void Profit(int C, int S){
System.out.println(S-C);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: def sample(n):
if n<200:
print(200-n)
elif n<400:
print(400-n)
elif n<500:
print(500-n)
else:
div=n//100
if div*100==n:
print(0)
else:
print((div+1)*100-n)
n=int(input())
sample(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
if(n <= 200){
cout<<200-n;
return;
}
if(n <= 400){
cout<<400-n;
return;
}
int ans = (100-n%100)%100;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans=0;
if(n <= 200){
ans = 200-n;
}
else if(n <= 400){
ans=400-n;
}
else{
ans = (100-n%100)%100;
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a set of positive integers. Find the maximum xor of a non-empty subset from the given set.The first line contains an integer N denoting the number of elements in the set.
The next line contains N integers denoting the elements of the set.
1 <= N <= 10^5
1 <= Elements of set <= 10^9Print the maximum subset Xor.Sample Input:
3
2 4 5
Sample Output:
7
Explanation:
Subset {2, 5} has the maximum xor
Sample Input:
3
9 8 5
Sample Output:
13
Explanation:
Subset {8, 5} has the maximum xor, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static final int INT_BITS = 32;
static int findMaxxor(int[] arr, int n){
int index = 0;
for (int i = INT_BITS - 1; i >= 0; i--)
{
int maxInd = index;
int maxEle = Integer.MIN_VALUE;
for (int j = index; j < n; j++) {
if ((arr[j] & (1 << i)) != 0 && arr[j] > maxEle)
{
maxEle = arr[j];
maxInd = j;
}
}
if (maxEle == -2147483648)
continue;
int temp = arr[index];
arr[index] = arr[maxInd];
arr[maxInd] = temp;
maxInd = index;
for (int j = 0; j < n; j++) {
if (j != maxInd && (arr[j] & (1 << i)) != 0)
arr[j] = arr[j] ^ arr[maxInd];
}
index++;
}
int res = 0;
for (int i = 0; i < n; i++)
res ^= arr[i];
return res;
}
public static void main (String[] args) throws Exception{
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(r);
int n = Integer.parseInt(br.readLine());
String[] sval = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(sval[i]);
System.out.print(findMaxxor(arr,n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a set of positive integers. Find the maximum xor of a non-empty subset from the given set.The first line contains an integer N denoting the number of elements in the set.
The next line contains N integers denoting the elements of the set.
1 <= N <= 10^5
1 <= Elements of set <= 10^9Print the maximum subset Xor.Sample Input:
3
2 4 5
Sample Output:
7
Explanation:
Subset {2, 5} has the maximum xor
Sample Input:
3
9 8 5
Sample Output:
13
Explanation:
Subset {8, 5} has the maximum xor, I have written this Solution Code: INT_BITS=32
def maxSubarrayXOR(set,n):
index = 0
for i in range(INT_BITS-1,-1,-1):
maxInd = index
maxEle = -2147483648
for j in range(index,n):
if ( (set[j] & (1 << i)) != 0
and set[j] > maxEle ):
maxEle = set[j]
maxInd = j
if (maxEle ==-2147483648):
continue
temp=set[index]
set[index]=set[maxInd]
set[maxInd]=temp
maxInd = index
for j in range(n):
if (j != maxInd and
(set[j] & (1 << i)) != 0):
set[j] = set[j] ^ set[maxInd]
index=index + 1
res = 0
for i in range(n):
res =res ^ set[i]
return res
n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
print (maxSubarrayXOR(arr,n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a set of positive integers. Find the maximum xor of a non-empty subset from the given set.The first line contains an integer N denoting the number of elements in the set.
The next line contains N integers denoting the elements of the set.
1 <= N <= 10^5
1 <= Elements of set <= 10^9Print the maximum subset Xor.Sample Input:
3
2 4 5
Sample Output:
7
Explanation:
Subset {2, 5} has the maximum xor
Sample Input:
3
9 8 5
Sample Output:
13
Explanation:
Subset {8, 5} has the maximum xor, 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 a[N];
int maxsubaXor(int n)
{
int index = 0;
for (int i = 31; i >= 0; i--)
{
int maxInd = index;
int maxEle = INT_MIN;
for (int j = index; j < n; j++)
{
if ( (a[j] & (1 << i)) != 0
&& a[j] > maxEle )
maxEle = a[j], maxInd = j;
}
if (maxEle == INT_MIN)
continue;
swap(a[index], a[maxInd]);
maxInd = index;
for (int j=0; j<n; j++)
{
if (j != maxInd &&
(a[j] & (1 << i)) != 0)
a[j] = a[j] ^ a[maxInd];
}
index++;
}
int res = 0;
for (int i = 0; i < n; i++)
res ^= a[i];
return res;
}
void solve(){
int n; cin >> n;
for(int i = 0; i < n; i++)
cin >> a[i];
cout << maxsubaXor(n) << 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: Mohan was so excited to visit his college NIT Arunachal Pradesh campus after COVID along with his friends Shreya and Anuj. The Warden allocated the hostel room for him but he was curious to know about the fact that how students are getting rooms.
The warden told him that those students whose roll no is divisible by 2 will get PAPUM HOSTEL and if it is odd then they will get "LOHIT HOSTEL". If the student is a girl, then she will get "Upper Wing" otherwise ‘Lower Wing’.The first line of input will contain a single integer t denoting the no. of test cases
Each test case contains only one line containing two space-separated inputs R for Roll No and S
for denoting Sex ‘B’ for boy and ‘G’ for girl
<b>Constraints</b>
1 ≤ t ≤ 10000
1 ≤ R ≤ 300Print the hostel name (LOHIT/PAPUM) with the upper wing (U) and lower wing (L).Sample Input
2
30 B
35 G
Sample Output
PAPUM L
LOHIT U
<b>Explanation:</b>
Here 30 is evenly divisible by two and B is the boy so he gets the Papum Hostel and is allocated a room in the lower wing., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int T = Integer.parseInt(st.nextToken());
while(T-->0){
st = new StringTokenizer(br.readLine());
int roomNo = Integer.parseInt(st.nextToken());
String gender = st.nextToken();
if(roomNo%2==0){
if(gender.equals("B")) System.out.println("PAPUM L");
else System.out.println("PAPUM U");
} else {
if(gender.equals("B")) System.out.println("LOHIT L");
else System.out.println("LOHIT U");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohan was so excited to visit his college NIT Arunachal Pradesh campus after COVID along with his friends Shreya and Anuj. The Warden allocated the hostel room for him but he was curious to know about the fact that how students are getting rooms.
The warden told him that those students whose roll no is divisible by 2 will get PAPUM HOSTEL and if it is odd then they will get "LOHIT HOSTEL". If the student is a girl, then she will get "Upper Wing" otherwise ‘Lower Wing’.The first line of input will contain a single integer t denoting the no. of test cases
Each test case contains only one line containing two space-separated inputs R for Roll No and S
for denoting Sex ‘B’ for boy and ‘G’ for girl
<b>Constraints</b>
1 ≤ t ≤ 10000
1 ≤ R ≤ 300Print the hostel name (LOHIT/PAPUM) with the upper wing (U) and lower wing (L).Sample Input
2
30 B
35 G
Sample Output
PAPUM L
LOHIT U
<b>Explanation:</b>
Here 30 is evenly divisible by two and B is the boy so he gets the Papum Hostel and is allocated a room in the lower wing., I have written this Solution Code: n = int(input())
for i in range(n):
a = input().split()
if int(a[0])%2==0:
print("PAPUM",end=" ")
else:
print("LOHIT",end = " ")
if a[1]=='G':
print("U")
else:
print("L"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohan was so excited to visit his college NIT Arunachal Pradesh campus after COVID along with his friends Shreya and Anuj. The Warden allocated the hostel room for him but he was curious to know about the fact that how students are getting rooms.
The warden told him that those students whose roll no is divisible by 2 will get PAPUM HOSTEL and if it is odd then they will get "LOHIT HOSTEL". If the student is a girl, then she will get "Upper Wing" otherwise ‘Lower Wing’.The first line of input will contain a single integer t denoting the no. of test cases
Each test case contains only one line containing two space-separated inputs R for Roll No and S
for denoting Sex ‘B’ for boy and ‘G’ for girl
<b>Constraints</b>
1 ≤ t ≤ 10000
1 ≤ R ≤ 300Print the hostel name (LOHIT/PAPUM) with the upper wing (U) and lower wing (L).Sample Input
2
30 B
35 G
Sample Output
PAPUM L
LOHIT U
<b>Explanation:</b>
Here 30 is evenly divisible by two and B is the boy so he gets the Papum Hostel and is allocated a room in the lower wing., I have written this Solution Code: #include <stdio.h>
int main(void) {
int t;
scanf("%d", &t);
while (t--) {
int roll;
char sex;
scanf("%d %c", &roll, &sex);
if (roll % 2)
printf("LOHIT %s\n", sex == 'B' ? "L" : "U");
else
printf("PAPUM %s\n", sex == 'B' ? "L" : "U");
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a <b>vertex cactus</b> with N nodes and M edges. You have to process Q queries on that graph, each query contains two nodes x and y. You have to find number of simple paths from x to y such that no node is visited twice in that path. Sir Isaac Newton himself has promised to give you chapo if you solve this. As the answer can be rather large, you should calculate it modulo 1000000007.
Note:- A connected undirected graph is called a vertex cactus, if each vertex of this graph belongs to at most one simple cycle.The first line contains two space-separated integers N and M
Next M lines contain two integers u and v denoting there is an edge between u and v.
The next line contains a single integer Q the number of queries
Next Q lines contain the two integers x and y, nodes on which you have to answer the query.
Constraints:
1 <= N, M, K <= 150000
1 <= x, y <= N
x != y
It is guaranteed that the given graph is a vertex cactus.Print the answer to each query modulo 1000000007 in a newline.Sample Input
10 11
1 2
2 3
3 4
1 4
3 5
5 6
8 6
8 7
7 6
7 9
9 10
4
1 2
6 9
9 2
9 10
Sample Output
2
2
4
1
Explanation:
For 1-2 paths are (1, 2), (1, 4, 3, 2)
For 6-9 paths are (6, 7, 9), (6, 8, 7, 9)
For 9-2 paths are (9, 7, 6, 5, 3, 2), (9, 7, 6, 5, 3, 4, 1, 2), (9, 7, 8, 6, 5, 3, 2), (9, 7, 8, 6, 5, 3, 4, 1, 2), 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
vector<int>NEB[sz];
int vis[sz];
int neww[sz],val[sz],is[sz];
int cnt=1;
set<int> NE[sz];
int dfs(int s,int f)
{
vis[s]=1;
for(auto it:NEB[s])
{ if(it==f) continue;
if(vis[it]==0)
{
val[s]+=dfs(it,s);
}else if(neww[it]==0)
{
val[s]=cnt;
neww[it]=-cnt;
is[cnt]=1;
cnt++;
}
}
//cout<<s<<" "<<val[s]<<" "<<neww[s]<<endl;
if(neww[s]<0)
{
neww[s]=val[s];
return 0;
}else if(val[s]==0)
{
neww[s]=cnt;
cnt++;
return 0;
}
neww[s]=val[s];
return neww[s];
}
int up[sz][20];
int tin[sz],tout[sz],ht[sz];
int tim=0;
int l=20;
int n,m;
void df(int s,int p,int h)
{
tim++;
//cout<<s<<endl;
tin[s]=tim;
up[s][0]=p;
if(s>n) h++;
ht[s]=h;
for(int i=1;i<l;i++)
{
up[s][i]=up[up[s][i-1]][i-1];
}
vis[s]=1;
for(auto it:NEB[s])
{
if(it!=p)
{
df(it,s,h);
}
}
tim++;
tout[s]=tim;
}
int isans(int u,int v)
{
if(tin[u]<=tin[v] && tout[u]>=tout[v]) return 1;
else return 0;
}
int lca(int u,int v)
{
if(isans(u,v)==1) return u;
else if(isans(v,u)==1) return v;
for(int i=l-1;i>=0;i--)
{
if(isans(up[u][i],v)!=1)
{
u=up[u][i];
}
}
return up[u][0];
}
vector<pii> X;
signed main()
{
fast
cin>>n>>m;
for(int i=0;i<m;i++)
{
int a,b;
cin>>a>>b;
NEB[a].pu(b);
NEB[b].pu(a);
X.pu(mp(a,b));
}
dfs(1,0);
memset(val,0,sizeof(val));
memset(vis,0,sizeof(vis));
// for(int i=1;i<=n;i++)
// {
// cout<<i<<" "<<" "<<neww[i]<<" "<<is[neww[i]]<<endl;
// }
for(int i=0;i<=n;i++)
{
NEB[i].resize(0);
}
for(int i=0;i<m;i++)
{
int a=X[i].fi;
int b=X[i].se;
int aa=neww[a];
int bb=neww[b];
if(aa!=bb)
{ NEB[a].pu(b);
NEB[b].pu(a);
}
}
for(int i=1;i<=n;i++)
{
if(is[neww[i]]==1)
{ int y=neww[i]+n+1;
NEB[y].pu(i);
NEB[i].pu(y);
}
}
df(1,1,0);
// for(int i=1;i<=n;i++)
// { cout<<i<<" "<<is[i]<<" ";
// for(auto it: NEB[i])
// {
// cout<<it<<" ";
// }cout<<endl;
// }
int k;
cin>>k;
int ans[sz];
ans[0]=1;
int mod=1000000007;
for(int i=1;i<=100000;i++)
{
ans[i]=(ans[i-1]*2LL)%mod;
}
while(k>0)
{
k--;
int a,b;
cin>>a>>b;
if(a==b)
{
cout<<1<<"\n";
}else
{
int c=lca(a,b);
int y=ht[a]+ht[b]-2*ht[c];
if(c>n ) y++;
cout<<ans[y]<<"\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Taro loves colours. Today she is interested in painting bricks. N bricks are arranged in a line. She needs to paint them using K colours. She does not want two adjacent bricks to be painted in same colour. Can you help her find the number of ways she can colour the given N bricks.
As the answer can be large print it modulo 1000000007.
Property: (a*b)%c = ((a%c)*(b%c))%c.The first and the only line of input contains 2 integers N and K.
Constraints
1 <= N <= 1000
2 <= K <= 1000Output a single integer, the number of ways in which she can colour the bricks modulo 1000000007.Sample Input
2 2
Sample Output
2
Explanation: Let the colours be 1 and 2. The ways of colouring the bricks are "1 2" or "2 1".
Sample Input:
1 10
Sample Output:
10, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static final int MOD = 1000000007;
static long modPow(long base, long exp, long mod) {
long x = 1;
while (exp > 0) {
if (exp % 2 == 1)
x = (x * base) % mod;
base = (base * base) % mod;
exp = exp / 2;
}
return x;
}
static long getColorPossibilities(long bricks, long colors) {
return (colors%MOD * modPow(colors-1, bricks-1, MOD))%MOD;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String inp[] = br.readLine().split(" ");
System.out.println(getColorPossibilities(Long.parseLong(inp[0]), Long.parseLong(inp[1])));
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Taro loves colours. Today she is interested in painting bricks. N bricks are arranged in a line. She needs to paint them using K colours. She does not want two adjacent bricks to be painted in same colour. Can you help her find the number of ways she can colour the given N bricks.
As the answer can be large print it modulo 1000000007.
Property: (a*b)%c = ((a%c)*(b%c))%c.The first and the only line of input contains 2 integers N and K.
Constraints
1 <= N <= 1000
2 <= K <= 1000Output a single integer, the number of ways in which she can colour the bricks modulo 1000000007.Sample Input
2 2
Sample Output
2
Explanation: Let the colours be 1 and 2. The ways of colouring the bricks are "1 2" or "2 1".
Sample Input:
1 10
Sample Output:
10, I have written this Solution Code: k= list(map(int,input().split()))
res = k[1]
for _ in range(1,k[0]):
res = (res*(k[1]-1))%1000000007
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Taro loves colours. Today she is interested in painting bricks. N bricks are arranged in a line. She needs to paint them using K colours. She does not want two adjacent bricks to be painted in same colour. Can you help her find the number of ways she can colour the given N bricks.
As the answer can be large print it modulo 1000000007.
Property: (a*b)%c = ((a%c)*(b%c))%c.The first and the only line of input contains 2 integers N and K.
Constraints
1 <= N <= 1000
2 <= K <= 1000Output a single integer, the number of ways in which she can colour the bricks modulo 1000000007.Sample Input
2 2
Sample Output
2
Explanation: Let the colours be 1 and 2. The ways of colouring the bricks are "1 2" or "2 1".
Sample Input:
1 10
Sample Output:
10, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n>>k;
long long mod=1000000007;
long long ans=k;
for(int i=0;i<n-1;i++){
ans*=k-1;
ans=ans%mod;
}
cout<<ans<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Solve the following problem for X from 1 to N :-
Find the number of ways to select a pair (i, j) such that i < j and i != X and j != X and Arr[i] = Arr[j].First line of input contains a single integer, N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers where ith integer is the answer when X = i.Sample Input
5
4 4 1 1 1
Sample Output
3 3 2 2 2
Explanation:
For X=1 we have (3, 4) (3, 5) (4, 5)
For X=2 we have (3, 4) (3, 5) (4, 5)
For X=3 we have (1, 2) (4, 5)
For X=4 we have (1, 2) (3, 5)
For X=5 we have (1, 2) (3, 4), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader sc=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(sc.readLine());
int a[]=new int[n];
int count[]=new int[n+1];
String[] temp1=sc.readLine().trim().split(" ");
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(temp1[i]);
count[a[i]]++;
}
int val=0,total=0;
for(int i=0;i<n;i++){
val=count[i];
total=total+(val)*(val-1)/2;
}
StringBuilder sb=new StringBuilder();
for(int j=0;j<n;j++){
int result=total-count[a[j]]+1;
sb.append(result+" ");
}
System.out.println(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Solve the following problem for X from 1 to N :-
Find the number of ways to select a pair (i, j) such that i < j and i != X and j != X and Arr[i] = Arr[j].First line of input contains a single integer, N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers where ith integer is the answer when X = i.Sample Input
5
4 4 1 1 1
Sample Output
3 3 2 2 2
Explanation:
For X=1 we have (3, 4) (3, 5) (4, 5)
For X=2 we have (3, 4) (3, 5) (4, 5)
For X=3 we have (1, 2) (4, 5)
For X=4 we have (1, 2) (3, 5)
For X=5 we have (1, 2) (3, 4), I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
d=dict.fromkeys([i for i in arr],0)
for i in arr:
d[i]+=1
ans=0
for value in d.values():
if value>=2:
ans+=((value*(value-1))//2)
for i in range(n):
if arr[i] in d:
f1=(d[arr[i]]*(d[arr[i]]-1))//2
ans-=f1
f2=((d[arr[i]]-1)*(d[arr[i]]-2))//2
ans+=f2
print(ans,end=" ")
ans-=f2
ans+=f1
else:
print(0,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N integers. Solve the following problem for X from 1 to N :-
Find the number of ways to select a pair (i, j) such that i < j and i != X and j != X and Arr[i] = Arr[j].First line of input contains a single integer, N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers where ith integer is the answer when X = i.Sample Input
5
4 4 1 1 1
Sample Output
3 3 2 2 2
Explanation:
For X=1 we have (3, 4) (3, 5) (4, 5)
For X=2 we have (3, 4) (3, 5) (4, 5)
For X=3 we have (1, 2) (4, 5)
For X=4 we have (1, 2) (3, 5)
For X=5 we have (1, 2) (3, 4), 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>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin >> n;
vector<int> a(n);
rep(i,n){
cin>>a[i];
}
map<int,int> b;
rep(i,n) b[a[i]]++;
ll ans=0,all=0;
for (auto p : b) {
ll v = p.second;
all+=v*(v-1)/2;
}
rep(i,n){
ans=all-b[a[i]]+1;
cout << ans << " ";
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, reverse each word in the sentenceA string containing multiple words
ex:- "Welcome to this Javascript Guide!"A string with all words reversed
ex:- "emocleW ot siht tpircsavaJ !ediuG"Sample input:-
"Welcome to this Javascript Guide!"
Sample output:-
"emocleW ot siht tpircsavaJ !ediuG"
Explanation:-
The first word is reversed from "Welcome" to "emocleW"
and similarly all other wrods are reversed but the order of the words is same, I have written this Solution Code: function reverseBySeparator(string, separator) {
return string.split(separator).reverse().join(separator);
}
function reverseWords (str){
const step1 = reverseBySeparator(str,"")
const step2 = reverseBySeparator(step1," ")
console.log(step2)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, reverse each word in the sentenceA string containing multiple words
ex:- "Welcome to this Javascript Guide!"A string with all words reversed
ex:- "emocleW ot siht tpircsavaJ !ediuG"Sample input:-
"Welcome to this Javascript Guide!"
Sample output:-
"emocleW ot siht tpircsavaJ !ediuG"
Explanation:-
The first word is reversed from "Welcome" to "emocleW"
and similarly all other wrods are reversed but the order of the words is same, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.Stack;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main(String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
String str = sc.readLine();
char[] charArray = str.toCharArray();
Stack<Character> st = new Stack<>();
for (int i = 0; i < charArray.length; i++) {
while (i < charArray.length && charArray[i] != ' ') {
st.push(charArray[i]);
i++;
}
while (!st.isEmpty()) {
System.out.print(st.pop());
}
System.out.print(" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given positive integers A and B.
Let us calculate A+B (in decimal). If it does not involve a carry, print <b>Easy</b>; if it does, print <b>Hard</b>.The input contains two space separated numbers:
A B
<b>Constraints</b>
A and B are integers.
1 ≤ A, B ≤ 10^18If the calculation does not involve a carry, print Easy; if it does, print Hard.<b>Sample Input 1</b>
229 390
<b>Sample Output 1</b>
Hard
<b>Sample Input 2</b>
123456789 9876543210
<b>Sample Output 2</b>
Easy, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
long long a,b;
cin >> a >> b;
while(a>0 && b>0){
if((a%10)+(b%10)>=10){cout << "Hard\n";return 0;}
a/=10;b/=10;
}
cout << "Easy\n";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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