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zhuklara [117]
3 years ago
11

Write a C++ nested for loop code to print out the following onthe screen1 2 3 4 5 67 8 91 2 3 4 5 67 81 2 3 4 5 671 2 3 4 561 2

3 451 2 341 23121
Computers and Technology
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer:

#include<iostream>

using namespace std;

//main function

int main(){

   

  //nested for loop

   for(int i=9;i>=1;i--){

       for(int j=1;j<=i;j++){

          cout<<j<<" ";  //display

       }

   }

   return 0;

}

Explanation:

Include the library iostream for using the input/output instruction in the c++ programming.

Create the main function and takes nested for loop. Nested for loop means, for loop inside the another for loop.

For every value of outside for loop, inside for loop execute.

we make outer for loop in decreasing format, means it start from 9 and goes to 1 and inside for loop make in increasing format, means loop start from 1 and it goes to that value which is provided by the outer loop.

and print the output for every cycle.

Lets dry run the code:

the outer loop starts from 9 and it checks the condition 9>=1, condition true. then the program moves to the inner loop which starts from 1 and goes to the 9.

first, it also checks the condition 1 <= 9, condition true and prints the number from 1 to 9.

then,  i is reduced by 1. it means i become 8.

then, the above process continues from 1 to 8 and so on...

the loop process will terminate if the outer loop terminate.

Finally, we get the output.

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Answer:

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int MaxMark(int* arr, int size) {

   int maxMark = 0;

   if (size > 0) {

       maxMark = arr[0];

   }

   for (int i = 0; i < size; i++) {

       if (arr[i] > maxMark) {

           maxMark = arr[i];

       }

   }

   return maxMark;

}

int SumMarks(int* arr, int size) {

   int sum = 0;

   for (int i = 0; i < size; i++) {

       sum += arr[i];

   }

   return sum;

}

float AvgMark(int* arr, int size) {

   int sum = SumMarks(arr, size);

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}

int main()

{

   int student0[] = { 7, 5, 6, 9 };

   int student1[] = { 3, 7, 7 };

   int student2[] = { 2, 8, 6, 1, 6 };

   int* marks[] = { student0, student1, student2 };

   int nrMarks[] = { 4, 3, 5 };

   int nrStudents = sizeof(marks) / sizeof(marks[0]);

   for (int student = 0; student < nrStudents; student++) {              

       printf("Student %d: max=%d, sum=%d, avg=%.1f\n",  

           student,

           MaxMark(marks[student], nrMarks[student]),

           SumMarks(marks[student], nrMarks[student]),

           AvgMark(marks[student], nrMarks[student]));

   }

   return 0;

}

Explanation:

Here is an example using a jagged array. Extend it to 15 students yourself. One weak spot is counting the number of marks, you have to keep it in sync with the array size. This is always a problem in C and would better be solved with a more dynamic data structure.

If you need the marks to be float, you can change the types.

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