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cupoosta [38]
3 years ago
5

Write a program that reads a list of integers into a vector, and outputs the two smallest integers in the list, in ascending ord

er. The input begins with an integer indicating the number of integers that follow. Ex: If the input is: 5 10 5 3 21 2 the output is: 2 and 3 You can assume that the list of integers will have at least 2 values. To achieve the above, first read the integers into a vector. Hint: Make sure to initialize the second smallest and smallest integers properly. Submit your .cpp code with this question's link. Output runs are not required.
Computers and Technology
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer:

In C++:

#include <bits/stdc++.h>

#include <iostream>

#include <vector>

using namespace std;

int main(){

vector<int> vectItems;

cout << "Vector length: ";

int ln; cin>>ln;

int num;

for (int ikk = 0; ikk < ln; ikk++){

 cin >> num;

 vectItems.push_back(num);}

int small, secsmall;

small = secsmall = INT_MAX;  

for (int ikk = 0; ikk < ln; ikk++){

 if(vectItems[ikk] < small){

     secsmall = small;  

           small = vectItems[ikk];   }

 else if (vectItems[ikk] < secsmall && vectItems[ikk] != small) {

           secsmall = vectItems[ikk];} }

 cout<<small<<" "<<secsmall;

 return 0;}

Explanation:

See attachment for program file where comments are used for explanation

Download cpp
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Motherboard

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3 years ago
Given two Strings String s1 = "11223351638791377189728193"; String s2 = "983763978669976862724569282405742578"; and String varia
inn [45]

Answer:

import java.math.BigInteger;  

public class Main {  

public static void main(String[] args)  {  

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

BigInteger num1 = new BigInteger(s1);  

BigInteger num2 = new BigInteger(s2);  

BigInteger product = num1.multiply(num2);  

String s3 = product.toString(10);

System.out.println(s3);  

}  

}  

Explanation:

Because s1 and s2 contain large integers, the program need the biginteger module in other to compute the products of s1 and s2 when converted to integers.

This line initializes the two string variables s1 and s2

String s1 = "11223351638791377189728193";   String s2 = "983763978669976862724569282405742578";

The next two lines convert the two strings to BigInteger. This allows the program to multiply large integers

<em> BigInteger a = new BigInteger(s1);   </em>

<em>BigInteger b = new BigInteger(s2);  </em>

<em />

This computes the product of the two big integers

BigInteger product = num1.multiply(num2);  

This converts the product to String s3

String s3 = product.toString(10);

This prints s3

System.out.println(s3);  

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7 0
3 years ago
Show the array that results from the following sequence of key insertions using a hashing system under the given conditions: 5,
sergejj [24]

Answer:

a) Linear probing is one of the hashing technique in which we insert values into the hash table indexes based on hash value.

Hash value of key can be calculated as :

H(key) = key % size ;

Here H(key) is the index where the value of key is stored in hash table.

----------

Given,

Keys to be inserted are : 5 , 205, 406,5205, 8205 ,307

and size of the array : 100.

First key to be inserted is : 5

So, H(5) = 5%100 = 5,

So, key 5 is inserted at 5th index of hash table.

-----

Next key to inserted is : 205

So, H(205) = 205%100 = 5 (here collision happens)

Recompute hash value as follows:

H(key) =(key+i) % size here i= 1,2,3...

So, H(205) =( 205+1)%100 = 206%100 = 6

So, key 205 is inserted in the 6th index of the hash table.

----------

Next Key to be inserted : 406

H(406) = 406%100 = 6(collision occurs)

H(406) =(406+1) %100 = 407%100 = 7

So, the value 406 is inserted in 7the index of the hash table.

-----------------

Next key : 5205

H(5205) = 5205%100 = 5(collision)

So, H(5205) = (5205+1)%100 = 6( again collision)

So, H(5205) = 5205+2)%100 = 7(again collision)

So, H(5205) = (5205+3)%100 = 8 ( no collision)

So, value 5205 is inserted at 8th index of the hash table.

-------------

Similarly 8205 is inserted at 9th index of the hash table because , we have collisions from 5th to 8th indexes.

-------------

Next key value is : 307

H(307) = 307%100 = 7(collision)

So, (307+3)%100 = 310%100 = 10(no collision)  

So, 307 is inserted at 10th index of the hash table.

So, hash table will look like this:

Key       index

5         5

205         6

406         7

5205 8

8205 9

307         10

b) Quadratic probing:

Quadratic probing is also similar to linear probing but the difference is in collision resolution. In linear probing in case of collision we use : H(key) = (key+i)%size but here we use H(key) =( key+i^2)%size.

Applying Quadratic probing on above keys:.

First key to be inserted : 5.

5 will go to index 5 of the hash table.

-------

Next key = 205 .

H(205) = 205%100 = 5(collision)

So. H(key)= (205+1^2)%100 = 6(no collision)

So, 205 is inserted into 6th index of the hash table.

--------

Next key to be inserted 406:

So, 406 %100 = 6 (collision)

(406+1^2)%100 = 7(no collision)

So, 406 is moved to 7th index of the hash table.

----------

Next key is : 5205

So, 5205%100 = 5 (collision)

So, (5205+1^2)%100 = 6 ( again collision)

So, (5205+2^2)%100 = 9 ( no collision)

So, 5205 inserted into 9th index of hash table.

-----------

Next key is 8205:

Here collision happens at 5the , 6the , 9th indexes,

So H(8205) = (8205+4^2)%100 = 8221%100 = 21

So, value 8205 is inserted in 21st index of hash table.

--------

Next value is 307.

Here there is collision at 7the index.

So, H(307) = (307+1^2)%100 = 308%100= 8.

So, 307 is inserted at 8the index of the hash table.

Key           Index

5                  5

205                  6

406                  7

5205                9

8205               21

307                   8

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Very few games have any sort of narrative, not even a simple one that involves setting the mood through color, sound, environmen
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Answer:

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Explanation:

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Answer:

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