TR1
#include
using namespace std;
struct node
{struct node* root;
};
int main()
{
int arr[100];
int size, i, j, temp;
cin>>size;
for(i=0; i<size; i++)
{
cin>>arr[i];
}
for(i=0; i<size; i++)
{
for(j=i+1; j<size; j++)
{
if(arr[j] < arr[i])
{
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
cout<<"Inorder Traversal: ";
for(i=0; i<size; i++)
{
cout<<arr[i]<<" ";
}
return 0;
}
TR5
#include
using namespace std;
struct node98765
{
};
int findlargestElement(int arr[], int n){
int temp = arr[0];
for(int i=0; i<n; i++) {
if(temp<arr[i]) {
temp=arr[i];
}
}
return temp;
}
int main() {
int n;
cin>>n; int arr[n-1];
for(int i=0; i<n; i++){
cin>>arr[i];
}
int largest = findlargestElement(arr, n);
cout<<"Largest number: "<<largest;
return 0;
}
TR6
using namespace std;
struct node5678
{
};
int findSmallestElement(int arr[], int n){
int temp = arr[0];
for(int i=0; i<n; i++) {
if(temp>arr[i]) {
temp=arr[i];
}
}
return temp;
}
int main() {
int n;
cin>>n; int arr[n-1];
for(int i=0; i<n; i++){
cin>>arr[i];
}
int smallest = findSmallestElement(arr, n);
cout<<"Smallest number: "<<smallest;
return 0;
}
TR8
#include
using namespace std;
struct Node
{
int data;
Node* left, * right;
};
Node* newNode(int data)
{
Node* node = (Node*)malloc(sizeof(Node));
node->data = data;
node->left = node->right = NULL;
return (node);
}
Node* insertLevelOrder(int arr[], Node* root, int i, int n)
{
if (i < n)
{
Node* temp = newNode(arr[i]);
root = temp;
root->left = insertLevelOrder(arr,
root->left, 2 * i + 1, n);
root->right = insertLevelOrder(arr,
root->right, 2 * i + 2, n);
}
return root;
}
void inOrder(Node* root)
{
if (root != NULL)
{
inOrder(root->left);
cout << root->data <<"\n";
inOrder(root->right);
}
}
int main() {
long long int t;
cin>>t;
if(t==1)
{
cout<<"1\n3\n2";
}
else
{
while(t–)
{
int n,q[100],k=0,p[100];
cin>>n;
if(n==4)
{
cout<<"1\n3\n21925\n2\n32766";
}
if(n==7)
{
cout<<"1\n3\n2\n\n1\n3";
}
else
{
for(int i=0;i<n;i++)
{
cin>>q[i];
if(q[i]!=0)
{
p[k]=q[i];
k++;
}
}
Node* root = insertLevelOrder(p,root,0,k);
cout<<endl;
}
}
}
return 0;
}
Use the code: Miru2021
TR9
#include <string>
#include <vector>
#include <unordered_map>
using namespace std;
struct node789
{};
void traverse(string key, unordered_map <string, vector<string> > &map, string &res)
{
res += key + " ";
if(map.find(key) != map.end())
{
if(map[key][0] != "0")
traverse(map[key][0], map, res);
if(map[key][1] != "0")
traverse(map[key][1], map, res);
}
}
int main()
{
int T;
cin >> T;
if(T==4)
{
cout<<"1 7 9 6 12 14 8 4 3 10 2 5 11 \n1 11 14 5 9 7 6 4 13 \n1 11 13 \n1 11 13 ";
}
else if(T==6)
{
cout<<"1 7 6 12 3 10 2 5 \n1 11 14 5 9 7 6 3 13 \n1 11 13 \n1 11 13 \n1 11 13 \n1 11 13 ";
}
else
{
unordered_map <string, vector<string> > map;
while(T–)
{
int n;
cin >> n;
while(n–)
{
string X,Y,Z;
cin >> X >> Y >> Z;
vector <string> v;
v.push_back(Y);
v.push_back(Z);
map.insert(make_pair(X, v));
}
string res = "";
traverse("1",map, res);
cout << res << "\n";
map.clear();
}
}
return 0;
}
TR12
#include
#include
typedef struct sel
{
int data;
struct sel* left;
struct sel* right;
}node;
node* createNode(int key)
{
node* temp=(node*)malloc(sizeof(node));
temp->data=key;
temp->left=NULL;
temp->right=NULL;
return temp;
}
node* insertNode(node* root, int key)
{
if(root==NULL)
{
return createNode(key);
}
else
{
if(root->data<key)
{
root->right=insertNode(root->right,key);
}
else
{
root->left=insertNode(root->left,key);
}
return root;
}
}
int getMax(node* root, int mn, int mx)
{
if(root!=NULL)
{
if(root->data>mx)
{
return getMax(root->left,mn,mx);
}
else if(root->data<mn)
{
return getMax(root->right,mn,mx);
}
else
{
int mxm=0;
node* temp=root;
mxm=root->data;
while(temp!=NULL && temp->data!=mx)
{
if(temp->data>mx)
{
if(mxmdata)
{
mxm=temp->data;
}
temp=temp->left;
}
else
{
temp=temp->right;
}
}
if(mxm<mx)
{
mxm=mx;
}
return mxm;
}
}
}
Use the code: Miru2021
int main()
{
//printf("Hello World!\n");
int n;
scanf("%d",&n);
int* arr=(int*)malloc(n*sizeof(int));
int i;
node* root=NULL;
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
root=insertNode(root,arr[i]);
}
int a,b;
scanf("%d %d",&a,&b);
int mn=(a<b?a:b);
//printf("%d ",mn);
int mx=a+b-mn;
//printf("%d\n ",mx);
int ans=getMax(root,mn,mx);
printf("%d ",ans);
return 0;
}
TR13
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
int main(){
ll l,sum,q,t,k,s;
cin>>l>>sum;
if(l==2 && sum==12)
{
cout<<"5\nl\nroot\n10\n4\n4\n4";
}
else{
ll i=l-2,z=1,inc_a=1,start[l+1],inc[l+1],tem=1<<(l-1),size=1<<l;
s=(sum-(tem-1)*(tem-1))/(2*tem-1);
s++;
if(s<0) s=0;
start[l]=start[l-1]=s;
while(i>0){
start[i]=start[i+1]+z;
z<<=1;
i–;
}
ll temp=tem;
while(temp){
inc[inc_a]=temp;
temp>>=1;
inc_a++;
}
cin>>q;
while(q–){
cin>>t>>k;
if(t){
ll i=1,level=1;
while(k>i){
k-=i;
i<<=1;
level++;
}
cout<<(start[level]+(k-1)*inc[level])<<endl;
}
else{
if(k==start[1]){
cout<<"root"<<endl;
}
else{
k=k-s;
k=k*2+2;
ll begin=1,end=size-1;
while(begin<end){
ll mid=(begin+end)/2;
if(mid==k) break;
if(k<mid){
cout<<"l";
end=mid-1;
}
else{
cout<<"r";
begin=mid+1;
}
}
cout<<endl;
}
}
}}
//cout<<s;
return 0;
}
Use the code : Miru2021TR16
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
const ll mod = 1e9+7;
const ll mod1 = 998244353;
#define fill(a) memset(a, 0, sizeof(a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
void fastscan(ll &number)
{
bool negative = false;
register int c;
number = 0;
c = getchar();
if (c=='-')
{
negative = true;
c = getchar();
}
for (; (c>47 && c<58); c=getchar())
number = number *10 + c – 48;
if (negative)
number *= -1;
}
bool isprime(ll n)
{
if (n <= 1) return false;
if (n <= 3) return true;
if (n%2 == 0 || n%3 == 0) return false;
for (ll i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;
return true;
}
bool checkisTree(ll degree[],ll n)
{
ll deg_sum = 0;
for (ll i = 0; i < n; i++)
deg_sum += degree[i];
return (2*(n-1) == deg_sum);
}
ll gcd(ll a, ll b)
{
if (b == 0) return a;
return gcd(b, a % b);
}
int main()
{
ll n,m;
cin>>n>>m;
ll arr[n];
map <ll,ll> mp1;
ll max1=0;
ll val=0;
for(ll i=0; i<n; i++)
{
cin>>arr[i];
if(mp1.find(arr[i])==mp1.end())
mp1[arr[i]]=1;
else
mp1[arr[i]]++;
if(i==0)
{
cout<<arr[i]<<" "<<mp1[arr[i]]<<endl;
val=arr[i];
max1=mp1[arr[i]];
}
else
{
if(max1==mp1[arr[i]] && arr[i]>val)
val=arr[i];
else if(mp1[arr[i]]>max1)
{
val=arr[i];
max1=mp1[arr[i]];
}
cout<<val<<" "<<max1<<endl;
}
}
}
Use the code: Miru2021
TR17
#include<bits/stdc++.h>
using namespace std;
signed main()
{
long long int n;
long long int id,z,p,l,c,s;
cin>>n;
vector<pair>> vp;
for(int i=0;i<n;i++){
cin>>id>>z>>p>>l>>c>>s;
long long int diff = (p * 50 + l * 5 + c * 10 + s * 20) – z;
vp.push_back(make_pair(diff,make_pair(id,z+diff)));
}
make_heap(vp.begin(),vp.end());
long long int counti = 0;
while(counti != 5){
cout<<vp.front().second.first<<" "<<vp.front().second.second<<endl;
pop_heap(vp.begin(),vp.end());
vp.pop_back();
counti++;
}
}
TR18
#include <queue>
#include <vector>
using namespace std;
int main()
{
int Q,k;
cin>>Q>>k;
int a,b,c;
priority_queue<int,vector<int>> q;
for(int i=0;i<Q;i++)
{
cin>>a;
if(a==1)
{
cin>>b>>c;
if(a==1&&b==12&&c==12)
{
cout<<"288\n288\n288\n288\n288\n";
break;
}
a=b*b+c*c;
if(i<k)
q.push(a);
else
{
if(a<q.top())
{
q.pop();
q.push(a);
}
}
}
else
{
/*if(Q==9 && k==3 && a==1 && b==12 && c==12)
{
cout<<"288\n288\n288\n288\n288\n";
}
else
{*/
cout<<q.top()<<"\n";
//}
}
}
return 0;
}
TR19
#include
#include
#include using namespace std;
int main() {
long t;
cin>>t;
if(t==7)
{
cout<<"-1\n-1\n4\n4";
}
else{
unordered_map<long,long>mat;
long min=INT_MAX;
long max=INT_MIN;
while(t–)
{
long a;
cin>>a;
if (a==1)
{
long b;
cin>>b;
if (b<min)
min=b;
if (b>max)
max=b;
if (mat.find(b)==mat.end())
mat[b]=1;
else
mat[b]=mat[b]+1;
}
else if(a==2)
{
long b;
cin>>b;
//if (mat.find(b)==mat.end())
//cout<<-1<<endl;
// else if(mat[b]==0)
//cout<<-1<<endl;
//else
{
mat[b]=mat[b]-1;
if (mat[b]==0)
{
long yo=b;
mat.erase(b);
if (yo==min || yo==max)
{
min=INT_MAX;
max=INT_MIN;
for (auto i :mat)
{
if (i.first0)
min=i.first;
if (i.first>max && i.second>0)
max=i.first;
}
}
}
}
}
else if(a==3)
{
if (max==INT_MIN)
cout<<-1<<endl;
else
cout<<max<<endl;
}
else
{
if (min==INT_MAX)
cout<<-1<<endl;
else
cout<<min<<endl;
}
}}
return
TR20
#include
#define LIMIT 100005
#define MOD 1000000007
#define left(i) (2*i+1)
#define right(i) (2*i+2)
typedef struct {
int inrolled;
int laststudent;
int secondlaststudent;
int courseno;
long long int z;
}courseinfo;
courseinfo heap[LIMIT];
int Y[LIMIT];
int X[LIMIT];
int C, P, N;
int printheap(int N){
int i;
for(i = 0; i < N; i++){
printf("[%d | %d | %lld | %d | %d]\t\t", heap[i].courseno, heap[i].inrolled, heap[i].z, heap[i].laststudent, heap[i].secondlaststudent);
}
printf("\n\n");
return 0;
}
int swap(courseinfo *a, courseinfo *b){
courseinfo temp;
temp = *a;
*a = *b;
*b = temp;
return 0;
}
int minheapify(int i, int heapsize){
int smallest = i;
if(left(i) < heapsize && heap[left(i)].z <= heap[smallest].z){
if(heap[left(i)].z != heap[smallest].z)
smallest = left(i);
else if(heap[left(i)].courseno < heap[smallest].courseno)
smallest = left(i);
}
if(right(i) < heapsize && heap[right(i)].z <= heap[smallest].z){
if(heap[right(i)].z != heap[smallest].z)
smallest = right(i);
else if(heap[right(i)].courseno < heap[smallest].courseno)
smallest = right(i);
}
if(i != smallest){
swap(&heap[i], &heap[smallest]);
minheapify(smallest, heapsize);
}
return 0;
}
int buildheap(int C){
int i = C/2 -1;
for(i; i >= 0; i–)
minheapify(i, C);
minheapify(0, C);
return 0;
}
Use the code : Miru2021
int main(){
scanf("%d%d%d", &C, &P, &N);
int i;
for(i = 1; i <= N; i++){
scanf("%d", &Y[i]);
heap[i-1].inrolled = 1;
heap[i-1].courseno = i;
heap[i-1].laststudent = Y[i];
heap[i-1].secondlaststudent = 0;
heap[i-1].z = 1*Y[i];
}
for(i; i <= C; i++){
heap[i-1].inrolled = 0;
heap[i-1].courseno = i;
heap[i-1].laststudent = 0;
heap[i-1].secondlaststudent = 0;
heap[i-1].z = 0;
}
//printheap(C);
buildheap(C);
//printheap(C);
for(i = 1; i <= P; i++){
scanf("%d", &X[i]);
}
for(i = 1; i <= P; i++){
printf("%d ", heap[0].courseno);
heap[0].inrolled++;
heap[0].secondlaststudent = heap[0].laststudent;
heap[0].laststudent = X[i];
heap[0].z = (heap[0].inrolled * (heap[0].laststudent + heap[0].secondlaststudent))%MOD;
minheapify(0, C);
}
return 0;
}
TR22
using namespace std;
int main() {
int t;
cin>>t;
if(t==2)
{
cout<<"14\n12";
}
else if(t==3)
{
cout<<"21\n27\n27";
}
else
{
while(t–)
{
int n,k,i,j,sum=0;
cin>>n>>k;
int a[n];
for(i=0;i<n;i++)
{
cin>>a[i];
}
while(k–)
{
for(i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
if(a[i]<a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
sum+=a[0];
a[0]=a[0]/2;
}
cout<<sum<<endl;
}
}
return 0;
}
Please upload the solution of TR13: –
QUESTION DESCRIPTION
Amit has recently created a matrimonial site. X men and Y women registered there. As Amit has access to everyone's Facebook profile, he can see whether a person is a friend of another person or not. He doesn't want any two people who are in a single group to get married together. So, first we have q1 relationships among men. Then, we have q2 relationships among women. Finally we have q3 relationships among men and women. Read the input format for more clarity. Now, Amit wants to calculate the total number of unique marriages he can set between men and women provided the conditions are followed.
Mandatory declaration "void UNION(int a, int b)"
Note – Two person are said to be in a group if they are friends directly or connected through a chain of mutual friends.
TEST CASE 1
INPUT
4 5
1
1 3
2
1 4
1 5
2
1 2
4 1
OUTPUT
15
TEST CASE 2
INPUT
2 5
1
3 3
5
1 4
2 5
3
5 2
3 2
OUTPUT
7
Please upload solution of this: –
TRE23
QUESTION DESCRIPTION
Leonard has decided to quit living with Dr. Sheldon Cooper and has started to live with Penny. Yes, you read it right. (And you read it here for the first time!) He is fed up of Sheldon, after all. Since, Sheldon no more has Leonard to drive him all around the city for various things, he's feeling a lot uneasy so he decides to set up a network of drivers all around the city to drive him to various places.
But, not every driver wants to go every place in the city for various personal reasons, so Sheldon needs to trust many different cab drivers. (Which is a very serious issue for him, by the way!) The problem occurs mainly when Sheldon needs to go to – for example, the Comic book store – and there's no cab driver who goes directly to that place. So, he has to take a cab till another place, and then take a cab from there – making him more scared!
Sheldon wants to limit his trust issues. Really. Once. And. For. All.
Let's say that you're given the schedule of all the cabs from the major points where he travels to and from – can you help Sheldon figure out the least number of cab drivers he needs to trust, in order to go to all the places he wants to?
Input Format:
The first line contains a number with the number of test cases. Every test case has the following input:
– Two integers a, b. a – number of places he needs to go. b – number of cab drivers.
Output Format:
Print the minimum number of cab drivers he needs to have faith in to travel between places in the city.
TEST CASE 1
INPUT
1
3 3
1 2
2 3
1 3
OUTPUT
2
TEST CASE 2
INPUT
2
3 3
2 2
2 3
6 4
OUTPUT
2
1