ch loop outputs all of the elements c int i; for( i = 0; i <= 20; ++i) %3D { cout << v.at(i) « endl;
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A: Given: A C-program code snippet is given. Goal: We have to find the output of the code program.
correct answer will be
int i;
for( i=0 ; i<=20; ++i)
{
cout<< vector.at(i) <<endl;
}
Explanation:
In order to print up to 20 we need to take greater than equal to symbol so that it can reach to 20.
In vectors we need to use vector keyword in defining the particular vectors.
Step by step
Solved in 2 steps
- 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;}Write a function that accepts a number, N, and a vector of numbers, V. The function will return two vectors which will make up any pairs of numbers in the vector that add together to be N. Do this with nested loops so the the inner loop will search the vector for the number N-V(n) == V(m). n and m are indices in the vector of numbers. Example A = [1,2,3,4,5,6, 7] Google(5, A) Return [1,2,3,4] and [4,3,2,1] being the pairs that sum to 5. Notice that each pair appears twice. Try to write code that does not do this.1. def f(x):"""Evaluates function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likeInput(s) to the functionReturns-------out : ndarrayFunction `f`, evaluated at point(s) x"""# your code here 2. def grad_f(x):"""Evaluates gradient of a function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likePoint(s) at which gradient should be avalueatedReturns-------out : ndarrayGradient of the function `f` evaluaed at point(s) x"""# your code here # TEST 1. f(x)assert f(0.) == 0.assert np.allclose(f(np.array([2.5, 7.5])), np.array([-1.25, 41.25])) assert np.allclose(f(np.arange(-10, 10, 1)), np.arange(-10, 10, 1)**2 - 15*np.sin(np.pi/3*np.arange(-10, 10, 1)))x_min = 1.33668375assert np.allclose(f(x_min), -12.9944407) # TEST 2. grad_f(x)assert np.allclose(grad_f(0.), -15.7079)assert isinstance(grad_f(0.), float) assert np.allclose(grad_f(0), - 5*np.pi*np.cos(0)), 'Gradient at point 0 is wrong'assert np.allclose(grad_f(np.arange(-10, 10, 1)), 2*np.arange(-10,…
- The program first reads integer cityCount from input, representing the number of pairs of inputs to be read. Each pair has a string and a character. One City object is created for each pair and added to vector cityList. Write the PrintSelectedCities() function in the SmallTowns class using Print() to output all the City objects with tour status equal to 'N'. Ex: If the input is: 4 Sundance N Adomstown N Hum Y Hanna Y then the output is: City: Sundance, Tour: N City: Adomstown, Tour: N Note: The vector has at least one element. #include <iostream>#include <vector>using namespace std; class City {public:void SetNameAndTour(string newName, char newTour); char GetTour() const; void Print() const;private:string name;char tour;}; void City::SetNameAndTour(string newName, char newTour) {name = newName;tour = newTour;} char City::GetTour() const { return tour;} void City::Print() const { cout << "City: " << name << ", Tour: " << tour <<…Write a similarprogram to compare two vectors. we should make 6 test : same length the first is longer, the second is longer, equal, the first greater , the second greater; try different symbols not only lettersIn PYTHON Write a function that prints multiple times a vector from its first element up to the next occurrence of an element having value equal to a provide number Each print stops at the i-th occurrence of the provided number Use while loop(s) Function Name:vec_multi_print_to_number Parameters:a_vector,a_value Return:none Example: Input:my_vector=[5,6,7,8,9,5,6,7,8,9] Invokefunction:print_up_to_value(my_vector,7) Output:[5,6,7,8,9,5,6,7]
- Write a program that reads a list of words. Then, the program outputs those words and their frequencies. The input begins with an integer indicating the number of words that follow. Ex: If the input is: 5 hey hi Mark hi mark the output is: hey 1 hi 2 Mark 1 hi 2 mark 1 Hint: Use two vectors, one for the strings, another for the frequencies. Your program must define and use the following function:int GetFrequencyOfWord(vector<string> wordsList, string currWord) #include <iostream>#include <vector> #include <string>using namespace std; /* Define your function here */ int main() {/* Type your code here */ return 0;} Please help me with this problem using c++.We initialize the parameters to all zero values and run the linear perceptron algorithm through these points in a particular order until convergence . The number of mistakes made on each point are shown in the table below . ( These points correspond to the data point in the plot above ) 00 = Note : You should be able to arrive at the answer without programming . What is the resulting offset parameter 00 ? Enter the numerical value for 00 : 0 Label Coordinates Perceptron mistakes - What is the resulting parameter ? ( Enter as a vector , e.g. type [ 0,1 ] if = [ 01 ] . ) -1 -1 -1 -1 -1 +1 +1 +1 +1 +1 ( 0,0 ) ( 2,0 ) ( 3,0 ) ( 0,2 ) ( 2,2 ) ( 5,1 ) ( 5,2 ) ( 2,4 ) ( 4,4 ) ( 5,5 ) 19 10 5 9 11 0 3 1 1 STANDARD NOTATIONAdd comments in 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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