Question d’entretien chez WebKul

enter any odd positive no . and do output as n=3 @@@ @@@ @@@***@ @@@ @@@ n=5 @@@@@ @@@@ @@@@@ @ @@@@@*****@ @@@@@ @ @@@@@ @@@@@

Réponses aux questions d'entretien

Utilisateur anonyme

20 avr. 2019

#include int main() { int i,j,n; //int n=5; printf("enter a number"); scanf("%d",&n); int dk=n/2+1; for(i=1;i<=n;i++) { for(j=1;j<=n;j++){printf("@");} //left side side for(j=1;j<=1;j++){printf(" ");} //spaces if((i==1) || (i==n)) //for first and last line of right side { for(j=1;j<=n;j++){printf("@");} } else if(i==dk){ //for midline of right side for(j=1;j<=n;j++){printf("*");} printf("@"); } else{ //for remaining lines of right side printf("@"); } printf("\n"); } return 0; }

3

Utilisateur anonyme

13 août 2019

#include int main(void) { // your code goes here int a=7; int d= a/2+1; for(int i=1;id) { printf(" "); } if(i==1 || i==a) { for(int k=1;k<=a;k++) { printf("@"); } } else if(i==d) { for(int m=1;m<=a;m++) { printf("*"); } printf("@"); } else { printf("@"); } printf("\n"); } return 0; }

1

Utilisateur anonyme

6 sept. 2019

n=int(input("enter the number")) z=int(n/2) print("@"*n,"@"*n) for i in range(0,z-1): print("@"*n,"@") print("@"*n,"*"*n,"@") for i in range(0,z-1): print("@"*n,"@"*n)

Utilisateur anonyme

24 avr. 2020

def fun(number): at_rate="@"*number for x in range(number): if x%2==0: continue if x==1: print(at_rate+" "+at_rate) else: print(at_rate+" "+"@") print(at_rate+"*"*number+"@") for x in reversed(range(number)): if x%2==0: continue if x==1: print(at_rate+" "+at_rate,end="") else: print(at_rate+" "+"@",end="") print("") fun(5)