nt main() { int n1=0,n2=1,n3,i,number; printf("Enter the number of elements:"); scanf("%d",&number); printf("\n%d %d",n1,n2);//printing 0 and 1 for(i=2;i
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- include<stdio.h>
- int main()
- {
- int n1=0,n2=1,n3,i,number;
- printf("Enter the number of elements:");
- scanf("%d",&number);
- printf("\n%d %d",n1,n2);//printing 0 and 1
- for(i=2;i<number;++i)//loop starts from 2 because 0 and 1 are already printed
- {
- n3=n1+n2;
- printf(" %d",n3);
- n1=n2;
- n2=n3;
- }
- return 0;
- } .
Output.
Step by step
Solved in 3 steps with 1 images
- #ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…#include #include #include #define SIZE 5 int nums [SIZE] = [5,0,6,1,2); int main() { int i; pid_t pid; pid= fork(); if (pid == 0) { for (i = 0; i 0) { wait (NULL); } for (i = 0; i < SIZE; i++) printf("PARENT: %d\n", nums [i] + 1); /* LINE Y */ return 0; From the above-given program, please state what will be the output at lines X and Y. Explanation in detail required for output at both lines X and Y.ypedef struct { char *args[10]; int n; int *a; } BOX; // ... int main(){ BOX a[100]; BOX *p; BOX x; BOX * b[10]; // ... }
- #include #include #include int main() { int i, j; printf("Columns | JIn"); for (i = 1; i< 4; ++i) { printf("Outer %6d\n", i); for (j = 0; j< i; ++j) { printf(" Inner%10d\n", j); } } /* heading of outer for loop */ /* heading of inner loop */ %3D return (0); } Create a new code modifying it to use "while loop" instead of the "for loop"int main(){ long long int total; long long int init; scanf("%lld %lld", &total, &init); getchar(); long long int max = init; long long int min = init; int i; for (i = 0; i < total; i++) { char op1 = '0'; char op2 = '0'; long long int num1 = 0; long long int num2 = 0; scanf("%c %lld %c %lld", &op1, &num1, &op2, &num2); getchar(); long long int maxr = max; long long int minr = min; if (op1 == '+') { long long int sum = max + num1; maxr = sum; minr = sum; long long int res = min + num1; if (res > maxr) { max = res; } if (res < minr) { minr = res; } } else { long long int sum = max * num1; maxr = sum; minr = sum; long long int res = min * num1;…(Electrical eng.) Write a program that specifies three one-dimensional arrays named current, resistance, and volts. Each array should be capable of holding 10 elements. Using a for loop, input values for the current and resistance arrays. The entries in the volts array should be the product of the corresponding values in the current and resistance arrays (sovolts[i]=current[i]resistance[i]). After all the data has been entered, display the following output, with the appropriate value under each column heading: CurrentResistance Volts
- char[ ] letters=new char[5];int x = 0;while(x < 10){ar[x]='a';x++;}int main{void main} { int myArray[100],numint,result; printf("How many numbers to be entered?"); scanf("%d"&numint); getNumbers(myArray,numint); printNumbers(myArray,numint); result=sumNumbers{myArray,numint}; printf("Sum of numbers entered=%d",result); return 0; } Based on this program, write down the function prototype for below. Write the function getNumber(), printNumbers() and sumNumbers().int main{void main} { int myArray[100],numint,result; printf("How many numbers to be entered?"); scanf("%d"&numint); getNumbers(myArray,numint); printNumbers(myArray,numint); result=sumNumbers{myArray,numint}; printf("Sum of numbers entered=%d",result); return 0; } Based on this program, write down the function prototype for function getNumber().
- #include using namespace std; int main() { int numbers[10]; int i; for (i=0;i>numbers[i]; int numbers2[10],j,k; for(i=0;ic) { max1=rep; c=top; } } cout<#include <iostream>#include<iomanip>using namespace std; int main(){ const int ROW_HOURS=6; //declare variables const int CL_PAYRATE=2; double total_payout=0; double hours[ROW_HOURS][CL_PAYRATE]={ //declare array {0,12.85}, {0,12.85}, {0,12.85}, {0,11.75}, {0,11.75}, {0,11.15} }; cout<<fixed; cout<<setprecision(2); //setprecision of double values upto 2 decimal point for(int i=0;i<6;i++) //using loop get input from user { cout<<"Enter the hours worked by employee "<<i+1<<": "; cin>>hours[i][0]; //store values into hours } for(int i=0;i<6;i++) { double payout=hours[i][0]*hours[i][1]; //calculate payout of emplyee total_payout=total_payout+payout;…be recor #include #include minutes #include limit on int func(int, int, int, int); main(){ srand(time(NULL)); int a, b, c, fNum; printf("Choose three different numbers between 0-39:"); scanf ("%d%d%d", &a, &b, &c); fNum = func (a, b, c, 25); printf("\nThe result: %d", fNum); } int func (int ul, int u2, int u3, int iter){ srand (time (NULL)); int n1=0, i=0, count=0; for (;iSEE MORE QUESTIONS