#include<iostream.h>#include<conio.h> void main(){ clrscr(); //to clear the screen int i,j,k,n; cout<<“How many lines?”; cin>>n; n*=2; for(i=0;i<n;i+=2) { for(j=n-2;j>i;j-=2) cout<<” “; for(k=0;k<=i;++k) cout<<“*”; cout<<“n”; } for(i=0;i<n;i+=2) { for(j=0;j<i;j+=2) cout<<” “; for(k=n-1;k>i;–k) cout<<“*”; cout<<“n”; } getch(); //to stop the screen}
The object oriented programming has been developed with a view to overcome the drawbacks of conventional programming approaches. The OOP approach is based on certain concepts that help it attain its goal of overcoming the drawbacks of shortcomings of conventional programming approaches. These general concepts of OOP are given below: Also Read: How to write and …
#include<stdio.h> #include<conio.h> void main() { int i,n,x,large=0; clrscr(); //to clear the screen printf(“How many numbers?”); scanf(“%d”,&n); for(i=0;i<n;++i) { printf(“nEnter number %d:”,i+1); scanf(“%d”,&x); if(x>large) large=x; } printf(“nnThe largest number is %d”,large); getch(); //to stop the screen }
#include<iostream.h>#include<conio.h> void main(){ clrscr(); //to clear the screen int i,n,x,large=0; cout<<“How many numbers?”; cin>>n; for(i=0;i<n;++i) { cout<<“nEnter number “<<i+1<<“:”; cin>>x; if(x>large) large=x; } cout<<“nnThe largest number is “<<large; getch();}
#include<stdio.h> #include<conio.h> void main() { int x,y,z; clrscr(); //to clear the screen printf(“XtYtZtXY+Z”); for(x=0;x<=1;++x) for(y=0;y<=1;++y) for(z=0;z<=1;++z) { if(x*y+z==2) printf(“nn%dt%dt%dt1”,x,y,z); else printf(“nn%dt%dt%dt%d”,x,y,z,x*y+z); } getch(); //to stop the screen }
#include<iostream.h> #include<conio.h> void main() { clrscr(); //to clear the screen int i,n,x,sum=0; cout<<“How many numbers?”; cin>>n; for(i=1;i<=n;++i) { cout<<“nEnter number “<<i<<“: “; cin>>x; sum+=x; } x=sum/n; cout<<“nnAverage=”<<x; getch(); //to stop the screen }
#include<stdio.h> #include<conio.h> void main() { int i,j,k,n; clrscr(); //to clear the screen printf(“How many lines:”); scanf(“%d”,&n); n*=2; printf(“n”); for(i=1;i<n;i+=2) { for(j=n-1;j>i;j-=2) printf(” “); for(k=1;k<=i;++k) { if(i==n-1) printf(“*”); else if(k==1||k==i) printf(“*”); else printf(” “); } printf(“n”); } getch(); //to stop the screen; }
#include<stdio.h> #include<conio.h> void main() { int i,j,k,n; clrscr(); //to clear the screen printf(“How many lines?”); scanf(“%d”,&n); n*=2; printf(“n”); for(i=0;i<n;i+=2) { for(j=1;j<i;j+=2) printf(” “); for(k=n-1;k>i;–k) printf(“*”); printf(“n”); } getch(); //to stop the screen }
