Question 6What is the output of:Answer savedint sumAll(int], int);Marked out of4.00P Flagint main() {int nums] = {1, 2, 3, 4, 5, 6};questioncout << "Total sum: "<< sumAll(nums, sizeof(nums)/sizeof(int)) << endl;return 0;}int sumAll(int arr), int length) {int sum;for(int i=0; i

The function sumAll() recieves an array and size of array as function parameters. The for loop…

What is the output of. int sumAll(int], int); int main() { int nums] = {1, 2, 3, 4, 5, 6}; cout << Total sum: "«sumAll(nums, sizeof(nums)/sizeof(int)) << endl; return 0; int sumAll(int arr], int length) { int sum; for(int i=0; i

What is the output of. int sumAll(int], int); int main() { int nums] = {1, 2, 3, 4, 5, 6}; cout << Total sum: "«sumAll(nums, sizeof(nums)/sizeof(int)) << endl; return 0; int sumAll(int arr], int length) { int sum; for(int i=0; i

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Array

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Given an integer array of positive single digit values such as:int a[] = {8,4,2,6,9};1) Write a recursive arrayToN function which returns the concatenation of all array values as an integer value.This function should accept all required data as parameters and return a long integer value.ExamplesIf the array is 8 4 2 6 9 the arrayToN function returns the integer 84269.If the array is 0 2 6 8 9 3 5 1 the arrayToN function returns the integer 2689351.
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In questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.
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haskell is the programming language
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Find error #include<bits/stdc++.h> using namespace std; class Solution{ public: bool isPossible(vector<int>ank,int n,int mid) { int count=0; for(int i=0;i<rank.size()) { int val= (-1 + sqrt(1+(8*mid)/rank[i]))/2; count+=val; } return count>=n; } int findMinTime(int N, vector<int>&A, int L){ int low=*min_element(A.end()),high=1000000; int ans=high; while(low<=high) { int mid=low+(high)/2; if(isPossible(A,mid)) { ans=mid; high=mid; } else low=mid; } return ans; } }; int main() { int t; cin>>t; while(t--) { int l; cin >> l; vector<int>arr(l); for(int i = 0; i < l; i++){ cin >> arr[i]; } Solution ob; int ans = ob.findMin(n, *arr, l); cout…
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briefly comment the code#include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX int source,V,E,time,visited[20],G[20][20];void DFS(int i){int j;visited[i]=1;printf(" %d->",i+1);for(j=0;j<V;j++){if(G[i][j]==1&&visited[j]==0)DFS(j);}}int main(){int i,j,v1,v2;printf("\t\t\tGraphs\n");printf("Enter the no of edges:");scanf("%d",&E);printf("Enter the no of vertices:");scanf("%d",&V);for(i=0;i<V;i++){for(j=0;j<V;j++)G[i][j]=0;}/* creating edges :P */for(i=0;i<E;i++){printf("Enter the edges (format: V1 V2) : ");scanf("%d%d",&v1,&v2);G[v1-1][v2-1]=1; } for(i=0;i<V;i++){for(j=0;j<V;j++)printf(" %d ",G[i][j]);printf("\n");}printf("Enter the source: ");scanf("%d",&source);DFS(source-1);return 0;}
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helllp