Matrix Addition and Mulitplication Is given here

 #include <stdio.h>

#define max 100

int main() {

    int arr1[max][max],arr2[max][max],arr3[max][max],arr4[max][max];

    int k,i,j,n;

    printf("Enter Row and Column of Matrix(both Should be equal to perform the Operation): ");

    scanf("%d",&n);

    printf("Enter Your Array 1 Element\n");

    for(i=0;i<n;i++)

    for(j=0;j<n;j++){

        printf("Enter %d Row %d Column: ",i,j);

        scanf("%d",&arr1[i][j]);

    }

    printf("Enter Your Array 2 Element\n");

    for(i=0;i<n;i++)

    for(j=0;j<n;j++){

        printf("Enter %d Row %d Column: ",i,j);

        scanf("%d",&arr2[i][j]);

    }

    printf("\nYour Matrix 1 is \n");

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            printf("%d ",arr1[i][j]);

        }

        printf("\n");

    }

    printf("Your Matrix 2 is \n");

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            printf("%d ",arr2[i][j]);

        }

        printf("\n");

    }

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            arr3[i][j]+=arr1[i][j]+arr2[i][j];

        }

    }

    printf("\nYour Addition of Matrix is \n");

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            printf("%d ",arr3[i][j]);

        }

        printf("\n");

    }

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            arr4[i][j]=0;

            for(k=0;k<n;k++)

            arr4[i][j]+=arr1[i][k]*arr2[k][j];

        }

    }

    printf("\nYour mutiplication of Matrix is \n");

    for(i=0;i<n;i++){

        for(j=0;j<n;j++){

            printf("%d ",arr4[i][j]);

        }

        printf("\n");

    }

    

    return 0;

}

Array Implementation of Stack (Using predefined Value)

// program to implement array

#include <stdio.h>

#include <stdlib.h>

#include <limits.h>

struct stack{

      int top;

      int capacity;

      int *array;

};

struct stack *create(int size){

      struct stack *s = (struct stack *)malloc(sizeof(struct stack));

      if(!s){

            printf("Not enough space\n");

            return NULL;

      }

      s->capacity = size;

      s->top = -1;

      s->array = malloc(s->capacity*sizeof(int));

      if(!s->array){

            printf("Not enough space\n");

            return NULL;

      }

      return s;

}

int isempty(struct stack *s){

      return (s->top==-1);

}

int isfull(struct stack *s){

      return (s->top==s->capacity-1);

}

int size(struct stack *s){

      return (s->top+1);

}

void push(struct stack *s,int data){

      if(isfull(s)){

            printf("stack overflow\n");

            return;

      }

      else

      s->array[++s->top]=data;

}

int pop(struct stack *s){

      if(isempty(s)){

            printf("Stack is underflow\n");

            return -1;

      }

      return s->array[s->top--];

}

int peek(struct stack *s){

      if(isempty(s)){

            printf("stack is Empty\n");

            return -1;

      }

      else

      return (s->array[s->top]);

}

int main()

{

      int capacity=15,i;

      struct stack *stk = create(capacity);

      for(i=0;i<capacity;i++){

            push(stk,i);

      }

      printf("Top element in stack %d \n",peek(stk));

      printf("Deleted element in stack %d \n",pop(stk));

      printf("Top element in stack %d \n",peek(stk));

      printf("No of Element in Stack %d \n",size(stk));

    return 0;

}

Program to find the last index of an element of array using recursive function.

 // coded by Vaibhav Yadav

#include <iostream>

#define ll long long

ll index(ll a[],ll size,ll x);

using namespace std;

int main() {

    ll n,i,x;

    cout<<"Enter Your Number of Elements in array: ";

    cin>>n;

    ll a[n];

    for(i=0;i<n;i++){

        cout<<"Enter Your Element no "<<i+1<<": ";

        cin>>a[i];

    }

    cout<<"Enter the element you want to search for: ";

    cin>>x;

    if(index(a,n,x)>=0)

    cout<<"Element Is found at position "<<index(a,n,x)<<endl;

    else

    cout<<"Element is not found"<<endl;

    return 0;

}

ll index(ll a[], ll size, ll x){

    if(a[size-1]==x)

    return size-1;

    else

    index(a,size-1,x);

}

Round G 2022 - Kick Start 2022 Walktober Problem (C Code)

// Below is the Solution of the Round G 2022 -kickstart 2022 :) 

// Contact me if any question persist in your mind, I'm ready to help you :) 

// HAPPY CODING

#include <stdio.h>

int main(){

    int t,count=0;

    scanf("%d",&t);

    while(t--){

        int M,N,P,k=0,i,j;

        scanf("%d%d%d",&M,&N,&P);

        int max[N],pl[N],n[N];

        for(i=0;i<N;i++)

        max[i]=0;

        for(i=0;i<M;i++){

            for(j=0;j<N;j++){

                scanf("%d",&n[j]);

                if((i+1)!=P){

                if(max[j]<n[j])

                max[j]=n[j];

                }

                else{

                    pl[j]=n[j];

                }

            }

        }

        int ans=0;

        for(i=0;i<N;i++){

            if(max[i]>pl[i])

            ans+=(max[i]-pl[i]);

            else

            ans+=0;

        }

        printf("Case #%d: %d\n",++count,ans);

    }

    return 0;

}

Problem Solution of GBus count, Coding Practice with Kick Start Session #3 - Kick Start 2022 in C Language.

 

Problem

There exist some cities that are built along a straight road. The cities are numbered 1,2,3,…$1,2,3,\dots$ from left to right.

There are N$N$ GBuses that operate along this road. For each GBus, the range of cities that it serves is provided: the i$i$-th gBus serves the cities with numbers between Ai${\mathbf{A}}_{\mathbf{i}}$ and Bi${\mathbf{B}}_{\mathbf{i}}$, inclusive.

We are interested in a particular subset of P$P$ cities. For each of those cities, we need to find out how many GBuses serve that particular city.

Input

The first line of the input gives the number of test cases, T$T$. Then, T$T$ test cases follow; each case is separated from the next by one blank line. (Notice that this is unusual for Kickstart data sets.)

In each test case:

• The first line contains one integer N$N$: the number of GBuses.
• The second line contains 2N$N$ integers representing the ranges of cities that the buses serve, in the form A1${\mathbf{A}}_{\mathbf{1}}$ B1${\mathbf{B}}_{\mathbf{1}}$ A2${\mathbf{A}}_{\mathbf{2}}$ B2${\mathbf{B}}_{\mathbf{2}}$ A3${\mathbf{A}}_{\mathbf{3}}$ B3${\mathbf{B}}_{\mathbf{3}}$ ... AN${\mathbf{A}}_{\mathbf{N}}$ BN${\mathbf{B}}_{\mathbf{N}}$. That is, the first GBus serves the cities numbered from A1${\mathbf{A}}_{\mathbf{1}}$ to B1${\mathbf{B}}_{\mathbf{1}}$ (inclusive), the second GBus serves the cities numbered from A2${\mathbf{A}}_{\mathbf{2}}$ to B2${\mathbf{B}}_{\mathbf{2}}$ (inclusive), and so on.
• The third line contains one integer P$P$: the number of cities we are interested in, as described above. (Note that this is not necessarily the same as the total number of cities in the problem, which is not given.)
• Finally, there are P$P$ more lines; the i$i$-th of these contains the number Ci${\mathbf{C}}_{\mathbf{i}}$ of a city we are interested in.

Output

For each test case, output one line containing Case #x$x$: y$y$, where x$x$ is the number of the test case (starting from 1), and y$y$ is a list of P$P$ integers, in which the i$i$-th integer is the number of GBuses that serve city Ci${\mathbf{C}}_{\mathbf{i}}$.

Limits

Memory limit: 1 GB.
1≤T≤10$1\le \mathbf{T}\le 10$.

Test Set 1

Time limit: 60 seconds.
1≤N≤50$1\le \mathbf{N}\le 50$.
1≤Ai≤500$1\le {\mathbf{A}}_{\mathbf{i}}\le 500$, for all i$i$.
1≤Bi≤500$1\le {\mathbf{B}}_{\mathbf{i}}\le 500$, for all i$i$.
1≤Ci≤500$1\le {\mathbf{C}}_{\mathbf{i}}\le 500$, for all i$i$.
1≤P≤50$1\le \mathbf{P}\le 50$.

Test Set 2

Time limit: 120 seconds.
1≤N≤500$1\le \mathbf{N}\le 500$.
1≤Ai≤5000$1\le {\mathbf{A}}_{\mathbf{i}}\le 5000$, for all i$i$.
1≤Bi≤5000$1\le {\mathbf{B}}_{\mathbf{i}}\le 5000$, for all i$i$.
1≤Ci≤5000$1\le {\mathbf{C}}_{\mathbf{i}}\le 5000$, for all i$i$.
1≤P≤500$1\le \mathbf{P}\le 500$.

Here is the solved code for this problem in c language:

#include<stdio.h>

int main()

{

    int t,ccase=0;

    scanf("%d",&t);

    while(t--)

    {

        int a,i,j,k;

        scanf("%d",&a);

        int c = 2*a;

        int b[c];

        for(i=0;i<c;i++)

        scanf("%d",&b[i]);

            

        int p;

        scanf("%d",&p);

        int q[p];

        int count[p];

        

        for(i=0;i<p;i++)

        count[i]=0;

        

        for(i=0;i<p;i++)

        scanf("%d",&q[i]);

        

        for(j=0;j<p;j++)

        {

        for(k=0;k<c;k++)

        {

            if((q[j] >= b[k]) && (q[j] <= b[k+1])) {

            count[j]++;

            }

            k+=1;

        }

      }

      printf("Case #%d: ",++ccase);

      for(i=0;i<p;i++)

      printf("%d ",count[i]);

      printf("\n");

     }

    return 0;

}