C program to accept customer details

Write a ‘C’ program to accept customer details such as: Account_no, Name, Balance in account, Assume Maximum 20 customers in the bank.
Write a function to print the account no and name of each customer with
balance below Rs.100.

Solution:-

#include<stdio.h>
#include<conio.h>
struct customer
{
int accno;
char cname[20];
int bal;
};
void disp(struct customer * , int);
void main()
{
struct customer c[20];
int i,n;
clrscr();
printf("\nHOW MANY CUSTOMER U WANT");
scanf("%d",&n);
printf("\nENTER THE ACCNO,CNAME,BAL");
for(i=0;i<n;i++)
{
scanf("%d%s%d",&c[i].accno,&c[i].cname,&c[i].bal);
}
disp(&c[0],n);

getch();
  }
  void disp(struct customer *d,int n)
  {
    int i;
    for(i=0;i<n;i++)
    {
if(d[i].bal<100)
{
printf("\n%d\t%s",d[i].accno,d[i].cname);
}
    }
 }

C program menu drive example

Write a menu driven program in ‘C’ that shows the working of a library.
The menu option should be
- Add book information.
- Display book information.
- List all books of given author.
- List the count of books in the library.
- Exit.
(Using Structure Array)

#include
#include
struct book
{
int bno;
char bname[20],author[20];
};
void main()
{
struct book b[10];
int i,n,m;
char t[10];
clrscr();
do
{
printf("\n1.ADD BOOK INFORMATION");
printf("\n2.DISPLAY BOOK INFORMATION");
printf("\n3.LIST ALL BOOKS OF GIVEN AUTHOR");
printf("\n4.TOTAL NO OF BOOKS IN LIBRARY");
printf("\n5.EXIT");
printf("\nENTER UR CHOICE");
scanf("%d",&m);
switch(m)
{
case 1:
printf("\nHOW MANY BOOKS WANT");
scanf("%d",&n);

printf("\nENTER THE BNO,BNAME & AUTHOR");
for(i=0;i {
scanf("%d%s%s",&b[i].bno,&b[i].bname,&b[i].author);
}
break;
case 2:
printf("\nBOOK DETAILS \n");
for(i=0;i {
printf("\n%d\t%s\t%s",b[i].bno,b[i].bname,b[i].author);
}
break;
case 3:
printf("\nENTER THE AUTHOR NAME");
scanf("%s",&t);
printf("\n BOOK LIST");
for(i=0;i {
if(strcmp(&b[i].author,&t)==0)
{
printf("\n BOOK NAME");
printf("\n%s",b[i].bname);
}
}
break;
case 4:
printf("\n%d",n);
break;
case 5:
exit(0);
}
}
while(m!=5);
getch();
}