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algol [13]
3 years ago
12

Write a while loop that adds the first 100 numbers (from 1 to 100) together. To do so, use a while loop that continues while the

condition of cur_num being less than or equal to stop_num is True. Inside the loop, it should add cur_num to the running total variable, and also add 1 to cur_num. Make sure to check at the end that the value of total reflects the total summation.

Computers and Technology
2 answers:
Anettt [7]3 years ago
6 0

Answer:

cur_num = 0

stop_num = 100

total = 0

while cur_num <= stop_num:

   total += cur_num

   cur_num += 1

print(total)

Explanation:

Initialize the variables

Initialize a while loop that iterates while cur_num is smaller than or equal to stop_num

Add the cur_num to total in each iteration to calculate the total

Increase the cur_num at then end of each iteration to control the loop

When the loop is done, print the total

Andreas93 [3]3 years ago
6 0

Answer:

I am writing JAVA and C++ program. Let me know if you want the program in some other programming language.  

Explanation:

JAVA program:

public class SumOfNums{

public static void main(String[] args) { //start of main function

int cur_num   = 1; // stores numbers

int stop_num= 100; // stops at this number which is 100

int total = 0; //stores the sum of 100 numbers

/* loop takes numbers one by one from cur_num variable and continues to add them until the stop_num that is 100 is reached*/

while (cur_num <= stop_num) {

 total += cur_num; //keeps adding the numbers from 1 to 100

//increments value of cur_num by 1 at each iteration

 cur_num = cur_num + 1;}

//displays the sum of first 100 numbers

System.out.println("The sum of first 100 numbers is: " + total); }}  

Output:

The sum of first 100 numbers is: 5050

C++ program

#include <iostream> //for input output functions

using namespace std; //to identify objects like cin cout

int main(){     //start of main() function body

int cur_num   = 1; // stores numbers

int stop_num= 100; // stops at this number which is 100

int total = 0;//stores the sum of 100 numbers

/* loop takes numbers one by one from cur_num variable and continues to add them until the stop_num that is 100 is reached*/

while (cur_num <= stop_num) {  

//keeps adding the numbers from 1 to 100

//increments value of cur_num by 1 at each iteration

 total += cur_num;

 cur_num = cur_num + 1; }  

//displays the sum of first 100 numbers

cout<<"The sum of first 100 numbers is: "<<total; }

The output is given in the attached screen shot.

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You are given a list of n positive integers a1, a2, . . . an and a positive integer t. Use dynamic programming to design an algo
Anna11 [10]

Answer:

See explaination for the program code

Explanation:

The code below

Pseudo-code:

//each item ai is used at most once

isSubsetSum(A[],n,t)//takes array of items of size n, and sum t

{

boolean subset[n+1][t+1];//creating a boolean mtraix

for i=1 to n+1

subset[i][1] = true; //initially setting all first column values as true

for i = 2 to t+1

subset[1][i] = false; //initialy setting all first row values as false

for i=2 to n

{

for j=2 to t

{

if(j<A[i-1])

subset[i][j] = subset[i-1][j];

if (j >= A[i-1])

subset[i][j] = subset[i-1][j] ||

subset[i - 1][j-set[i-1]];

}

}

//returns true if there is a subset with given sum t

//other wise returns false

return subset[n][t];

}

Recurrence relation:

T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times

T(1)=1

solving recurrence:

T(n)=T(n-1)+t

T(n)=T(n-2)+t+t

T(n)=T(n-2)+2t

T(n)=T(n-3)+3t

,,

,

T(n)=T(n-n-1)+(n-1)t

T(n)=T(1)+(n-1)t

T(n)=1+(n-1)t = O(nt)

//so complexity is :O(nt)//where n is number of element, t is given sum

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