The output of the following program is to draw a group of stars in the form of #include void main(){ int ij; for (i-0;i<5;i++}{ for (j=10;j>0;j-X if(j>5+il lj<5-i) cout<<" "; else cout<<"*";} cout<
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- int func(int a, int b) { return (aنقطة واحدة Let A = {a; b; c; d} and R= {(a; a); (b; c); (c; b); (d; d)} then R is Transitive Equivalent not transitivestruct ID { int x; int y; }; struct name { ID r; }; int main() { name a[2]; for (int i=0;i<2;i++) { a[i).r.x=i; a[i].r.y=i+1; } for(int i=0;i<2;i++) cout<#include<stdio.h> #include<stdarg.h> void fun1(int num, ...); void fun2(int num, ...); int main() { fun1(1, "Apple", "Boys", "Cats", "Dogs"); fun2(2, 12, 13, 14); return 0; } void fun1(int num, ...) { char *str; va_list ptr; va_start(ptr, num); str = va_arg(ptr, char *); printf("%s ", str); } void fun2(int num, ...) { va_list ptr; va_start(ptr, num); num = va_arg(ptr, int); printf("%d", num); }.#include using namespace std; int main() { int type; double Vth, Vg, Vd, Vs; cin>>type>>Vth>>Vg>>Vd>>Vs; return 0; }class A Q5: Correct the following code fragment and what will be the final results of the variable a and b { protected int x1,y1,z; public: A(a, b,c):x1(a+2),y1(b-1),z(c+2) { for(i=0; i<5;i++) x1++; y1++;z++;}}; class B { protected: int x,y; public: B(a,b):x(a+1),y(b+2) { for(i=0; i<5;i++) x+=2; y+=1;}}; class D:public B, virtual public A { private: int a,b; public: D(k,m,n): a(k+n), B(k,m),b(n+2),A(k,m,n) { a=a+1;b=b+1;}}; int main() {D ob(4,2,5);}#include using namespace std; (а) class Fraction{ int x, y; public: Fraction (int a, int b) {x=a; y=b;}; int getx(){return x;} int gety() {return y;} } ; int main() Fraction n(3, 7); ++n; cout << "x/y: " <« n.getx()<< "/" « n.gety()<c++ coding language I need help with part B and C please. If you are unable to do both, then PLEASE prioritize part C. I am really stuck and really can use the help. This is the code for c that was provided in order to guide me: const int N =31; // N parking spaces bool parking[N]; // the garage void EmptyTheLot(bool parking[], int N) { for(int i=0; i<N; i++) p[i]=false; // empty space } // returns -1 if no space found, //otherwise it returns 0<=i<N for a valid space. int FindSpace(int PlateNumber, bool parking[], int N) { // ????? } main() { EmptyTheLot(parking, N); // start with an empty parking garage. // get plate numbers and fill lot. }int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;Term by CodeChum Admin (JAVA CODE) Construct a class called Term. It is going to represent a term in polynomial expression. It has an integer coefficient and an exponent. In this case, there is only 1 independent variable that is 'x'. There should be two operations for the Term: public Term times(Term t) - multiplies the term with another term and returns the result public String toString() - prints the coefficient followed by "x^" and appended by the exponent. But with the following additional rules: if the coefficient is 1, then it is not printed. if the exponent is 1, then it is not printed ( the caret is not printed as well) if the exponent is 0, then only the coefficient is printed. Input The first line contains the coefficient and the exponent of the first term. The second line contains the coefficient and the exponent of the second term. 1·1 4·3 Output Display the resulting product for each of the test case. 4x^4#include using namespace std; class fraction{ float d,s; public: fraction() { d=0; s=1; fraction(int a, int b) { d=%3; s=b3; } void print() { cout<C Programming Language (Code With C Programming Language) Problem Title : Visible Trees There is a legend about a magical park with N × N trees. The trees are positioned in a square grid with N rows (numbered from 1 to N from north to south) and N columns (numbered from 1 to N from west to east). The height (in metres) of each tree is an integer between 1 and N × N, inclusive. Magically, the height of all trees is unique. Bunga is standing on the northmost point of the park and wants to count the number of visible trees for each Column. Similarly, Lestari is standing on the westmost point of the park and wants to count the number of visible trees for each Row. A tree X is visible if all other trees in front of the tree X are shorter than the tree X. For example, let N = 3 and the height (in metres) of the trees are as follows6 1 87 5 32 9 4 On the first column, Bunga can see two trees, as the tree on the third row is obstructed by the other trees. On the second column, Bunga can see…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
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