C language interview questions solution for freshers beginners placement tricky good pointers answers explanation operators data types arrays structures functions recursion preprocessors looping file handling strings switch case if else printf advance linux objective mcq faq online written test prime numbers Armstrong Fibonacci series factorial palindrome code programs examples on c++ tutorials and pdf
lets work out this with example now suppose for n=1 1<1 does not satisfy for n=2 1<2 satisfies 2%1==0 satisfies sum=0+1=1 i++(so i is 2 now) 2<2 does not satisfy for n=3 while 2<3 satisfies 3%2==0 does not satisfy for n=4 while 2<4 satisfies 2%4==0 satisfies sum=1+2=3 i=3(i++ i.e i=2+1=3) for n=5 while 3<5 satisfies 5%3==0 does not satisfy for n=6 while 3<6 satisfies 6%3==0 satisfies sum=3+3=6 So here Sum==n satisfies so 6 is a perfect number... Rocked it everyone can easily understand now
/* reverse of the value in using the program*/ #include #include void main() { int a[15][15],n,i,j; printf("Entr the number of the n=>"); scanf("%d",&n); printf("Enter the number you want to insert=>"); for(i=0;i0;i--) { for(j=n;j>0;j--) { printf("%d",a[i][j]); } getch(); } } ____________________________________________________________________________________________________________________ output _______ inter the value of n=2 value= 42,43,44,46; output is=46,44,43,42 is reverse order;
In code2 when i declared i,sum values at the starting it is not showing the results do u know the reason or there is any rule for for loop to give values in that loop
You can reduce the number of iterations by initializing 'sum with 1 and i with 2'. And why to check 'n%i == 0' upto 'i < n'. Check only upto 'i <= n/2'.
Excuse me, 6 is perfect number because 1+2+3 = 1*2*3, BUT with number 28 when i calculate : 1+2+4+7+14 = 28 and 1*2*4*7*14 = 784 <=== the sum of left doesn't equal with the multiply of right like 6=1+2+3=1*2*3 ???? I need to find all numbers include >=2 digits which have the sum = the multiplication, i wonder is it perfect number ? and how to find it ? thanks !!
#include main() { int q[10]={0},i,front=-1,rear=-1,max=10,n,item; printf("\n\tMENU\n1.ENQUEUE\n2.DEQUEUE\n3.DISPLAY\n4.EXIT\n"); do { printf("\nEnter your choice\n"); scanf("%d",&n); switch(n) { case 1: if(rear=0) { printf("The deleted item =%d",q[front]); if(front==rear) { front=-1; rear=-1; } else front++; } else printf("Underflow\n"); break;
case 3: if((front==-1)&&(rear==-1)) printf("The queue is empty\n"); else { printf("The elements of the queue are :"); for(i=front;i<=rear;i++) printf("%d\t",q[i]); } break;
Deleting a Node from the Linked List (assuming values in Data are unique)
Function Delete(Value)
Pointer : Found 'Pointer for finding required Node to Delete
Found = NULL 'Assuming required Node not found
IF there are no Nodes in the List then WRITE error message that can't delete a Node
ELSE Go to memory location pointed by Start pointer 'Begin at the Start (first Node) Start Loop and run until Next points to NULL 'Check each Node until Next pointer points to NULL
IF value in Data of current Node is equal to Value then 'If we find our required Node Found = current Node 'Mark the current Node with Found pointer
Exit Loop 'No need to finish the loop
ENDIF
Go to Node pointed by Next pointer of current Node
End Loop
IF the required Node is the first Node in the List then Point the Start pointer to NULL 'Node deleted
ELSE
Point Next of previous Node to the Node pointed by Next of Found 'Node deleted (Previous.Next = Found.Next) ENDIF ENDIF End Function
binary search Begin ACCEPT N Elements READ Element X to search
IF (range contains only one element): Look for desired value IF found WRITE Element X found
Else:
1. Get midpoint of range(N) 2. Determine which half of range contains the value 3. Search that half of the range using binary search Look for desired value IF found WRITE Element X found ENDIF ENDIF End
/*proram to calculate the pay of a worker*/ #include #include float pays(int hours,float rate); int main() { int hours,itrvar,num; float rate; float pay=0,totalpay=0;
printf("enter the number of workers"); scanf("%d",&num);
for(itrvar=0;itrvar0)/*condition to check hourly rate should be greater than 0*/ { pay=pays(hours,rate);/*call of pays function*/ printf("Pay at %f rupees/hr for %d is %f\n",rate,hours,pay); totalpay=totalpay+pay; }/*end of if*/ else { printf("Total pay is=%f",totalpay); exit(0); }/*end of else*/ }/*end of loop*/
printf("Total pay is=%f",totalpay); }/*end of main()*/ float pays(int hours,float rate) { float pay; if(hours<40)/*check for number of hours worked for calculating pay*/ pay=hours*rate; else if(hours>=40&&hours<60) pay=hours*1.5*rate; else if(hours>=60) pay=hours*2.0*rate; return pay; }/*end of pays()*/
/*proram to calculate the pay of a worker*/ #include #include float pays(int hours,float rate); int main() { int hours,itrvar,num; float rate; float pay=0,totalpay=0;
printf("enter the number of workers"); scanf("%d",&num);
for(itrvar=0;itrvar0)/*condition to check hourly rate should be greater than 0*/ { pay=pays(hours,rate);/*call of pays function*/ printf("Pay at %f rupees/hr for %d is %f\n",rate,hours,pay); totalpay=totalpay+pay; }/*end of if*/ else { printf("Total pay is=%f",totalpay); exit(0); }/*end of else*/ }/*end of loop*/
printf("Total pay is=%f",totalpay); }/*end of main()*/ float pays(int hours,float rate) { float pay; if(hours<40)/*check for number of hours worked for calculating pay*/ pay=hours*rate; else if(hours>=40&&hours<60) pay=hours*1.5*rate; else if(hours>=60) pay=hours*2.0*rate; return pay; }/*end of pays()*/
int SearchByName(char* Name, char* Value, int n) { int loop;
for(loop=0; loop4) printf("\n Illegal choice!!");
else {
switch(Choice) { case 1: printf("\n Enter name to search: "); /* Searching by name */ scanf("%s",Value); temp1= SearchByName(Name, Value, Num); if(temp1!=-1) printf("\n Name: %s\n ID: %d\n Marks: %d\n\n", Name[temp1], ID[temp1], Marks[temp1]); else printf("\n Name not found"); break;
case 2: printf("\n Enter ID to search: "); /* Searching by ID */ scanf("%d", &temp2); temp1= Search(ID, temp2, Num); if(temp1!=-1) printf("\n Name: %s\n ID: %d\n Marks: %d\n\n", Name[temp1], ID[temp1], Marks[temp1]); else printf("\n ID not found"); break;
case 3: printf("\n Students record with rank\n\n"); /* Printind data rankwise */ for(loop=0; loop<Num; loop++) printf("\n Rank: %d\n Name: %s\n ID: %d\n Marks: %d\n\n", loop+1, Name[loop], ID[loop], Marks[loop]); break;
case 4: printf("\n Enter ID to search: "); /* Searching name by ID */ scanf("%d", &temp2); temp1= Search(ID, temp2, Num); if(temp1!=-1) printf("\n Name: %s\n\n",Name[temp1]); else printf("\n ID not found"); break; } }
I tried code 2 using for loop instead of while loop, but it is not working. What could be the possible reason. I replaced i=1; sum = 0;
while(i<n){ if(n%i==0) sum=sum+i; i++; } by for(i=1;i<n;++i) { if(n%i==0) sum+=i; } The compiler shows no errors but blank space after the line "The perfect numbers are:". Please reply.
#include int is_perfect(int); int main() { int x; printf("Enter any number\n"); scanf("%d",&x); if(is_perfect(x)) printf("perfect\n"); else printf("not perfect");
74 comments:
thank u for ur program.
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lets work out this with example now suppose
for n=1
1<1 does not satisfy
for n=2
1<2 satisfies
2%1==0 satisfies
sum=0+1=1
i++(so i is 2 now)
2<2 does not satisfy
for n=3
while 2<3 satisfies
3%2==0 does not satisfy
for n=4
while 2<4 satisfies
2%4==0 satisfies
sum=1+2=3
i=3(i++ i.e i=2+1=3)
for n=5
while 3<5 satisfies
5%3==0 does not satisfy
for n=6
while 3<6 satisfies
6%3==0 satisfies
sum=3+3=6
So here Sum==n satisfies so 6 is a perfect number...
Rocked it everyone can easily understand now
It would be better to initialize Sum with 1 and i with 2.
brilliant
thanks
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give the answer for question --------------10. Write a c program to add two numbers without using addition operator.???
Thank you very much. This tutorial have helped me complete my university Assignments and Lab reports in time.
Thank u.....
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Parveen Arora
really i am very much thankful to u...its very much helpful to me..thanks a lot.
sir should we have to take the sum of values of i for the perfect no...
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nice concept used
/* reverse of the value in using the program*/
#include
#include
void main()
{
int a[15][15],n,i,j;
printf("Entr the number of the n=>");
scanf("%d",&n);
printf("Enter the number you want to insert=>");
for(i=0;i0;i--)
{
for(j=n;j>0;j--)
{
printf("%d",a[i][j]);
}
getch();
}
}
____________________________________________________________________________________________________________________
output
_______
inter the value of n=2
value= 42,43,44,46;
output is=46,44,43,42 is reverse order;
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Nice one raj
nice
In code 2,how about is "time complexity" ?
Thanks.
In code2 when i declared i,sum values at the starting it is not showing the results do u know the reason or there is any rule for for loop to give values in that loop
No time complexity
i wish allah will make yr dreams true
Code 1.
You can reduce the number of iterations by initializing 'sum with 1 and i with 2'.
And why to check 'n%i == 0' upto 'i < n'. Check only upto 'i <= n/2'.
Excuse me, 6 is perfect number because 1+2+3 = 1*2*3, BUT with number 28 when i calculate : 1+2+4+7+14 = 28 and 1*2*4*7*14 = 784 <=== the sum of left doesn't equal with the multiply of right like 6=1+2+3=1*2*3 ????
I need to find all numbers include >=2 digits which have the sum = the multiplication, i wonder is it perfect number ? and how to find it ?
thanks !!
This site is realy very good
i love your website thank you for your help
thanks a lot. :)
READ POL_TYPE,term,amount;
IF(pol_type is endownment)
Printf(“2. Money back”);
Printf(“3. Term assurance”)
If (term>=25){
Bonus=(amount/1000)*60;
WRITE BONUS
ENDIF
If (term>=20 && term<25)
Bonus=(amount/1000)*55;
WRITE BONUS
ENDIF
If (term>=10 && term<20)
Bonus=(amount/1000)*50;
WRITE BONUS
ENDIF
If (term<10){
Bonus=(amount/1000)*45;
WRITE BONUS
ENDIF
ENDIF
IF(pol_type is Money back)
If (term>=25)
Bonus=(amount/1000)*55;
WRITE BONUS
ENDIF
If (term>=20 && term<25)
Bonus=(amount/1000)*50;
WRITE BONUS
ENDIF
If (term>=10 && term<25)
Bonus=(amount/1000)*45;
WRITE BONUS
ENDIF
If (term<10){
Bonus=(amount/1000)*45;
IF(pol_type is Term assurance)
WRITE ““bonus not applicable”)
END
#include
main()
{
int q[10]={0},i,front=-1,rear=-1,max=10,n,item;
printf("\n\tMENU\n1.ENQUEUE\n2.DEQUEUE\n3.DISPLAY\n4.EXIT\n");
do
{
printf("\nEnter your choice\n");
scanf("%d",&n);
switch(n)
{
case 1:
if(rear=0)
{
printf("The deleted item =%d",q[front]);
if(front==rear)
{
front=-1;
rear=-1;
}
else
front++;
}
else
printf("Underflow\n");
break;
case 3:
if((front==-1)&&(rear==-1))
printf("The queue is empty\n");
else
{
printf("The elements of the queue are :");
for(i=front;i<=rear;i++)
printf("%d\t",q[i]);
}
break;
case 4:
break;
default:
printf("Invalid choice\n");
break;
}
}
while(n!=4);
}
#include
#include
int main(void)
{
int salesperson[10][2];
int index,totalsales=0;
float minsales,incentiveamount,averagesales=0;
for(index=0;index<10;index++)
{
printf("enter the salesperson id and his sales: \n");
scanf("%d%d",&salesperson[index][0],&salesperson[index][1]);
totalsales+=salesperson[index][1];
}
averagesales=(float)totalsales/(float)10;
minsales=(averagesales*2)/(float)3;
printf("\n Salesperson ID \t Incentive Amount \n");
printf("\n_____________________________________\n");
for(index=0;index<10;index++)
{
if(salesperson[index][1]>minsales)
{
incentiveamount=(float)(salesperson[index][1]-minsales)*(float)0.2;
printf("%d \t \t\t %f \n", salesperson[index][0],incentiveamount);
}
}
getch();
return 0;
}
pseudocode for Queue
Begin
Accept the N elements
INSERT Operation(insertion in stack)(to back)
Repeat for N-1
Insert the first element to Queue[0]
Insert the second element to Queue[1]
"
"
"
Insert the last element to Queue[N-1] "
REMOVE Operation(deletion from stack)(from front)
Repeat for N-1
REMOVE the first element from Queue[N-1]
REMOVE the second element from Queue[N-2]
"
"
"
REMOVE the last element from Queue[0] "
End
pseudocode for Queue(lab3_3.4)
Begin
Accept the N elements
INSERT Operation(insertion in stack)(to back)
Repeat for N-1
Insert the first element to Queue[0]
Insert the second element to Queue[1]
"
"
"
Insert the last element to Queue[N-1] "
REMOVE Operation(deletion from Queue)(from front)
Repeat for N-1
REMOVE the first element from Queue[N-1]
REMOVE the second element from Queue[N-2]
"
"
"
REMOVE the last element from Queue[0] "
DISPLAY Operation(Display Queue Elements
Repeat for N-1
WRITE the first element from Queue[0]
WRITE the second element from Queue[1]
"
"
"
WRITE the last element from Queue[N-1] "
No. of Element in queue
Repeat for NULL(iteration until the element stored in a queue reaches to null value)
Count++
END of Loop
WRITE Count
End
stack
pseudocode for Stack
Begin
Accept the N elements
PUSH Operation(insertion in stack)
Repeat for N-1
PUSH the first element to Stack[0]
PUSH the second element to Stack[1]
"
"
"
PUSH the last element to Stack[N-1] "
POP Operation(deletion from stack)
Repeat for N-1 to 0
POP the last element from Stack[N-1]
POP the second_last element from Stack[N-2]
"
"
"
POP the first element from Stack[0] "
End
linked list
pseudocode for LINKED LIST
Deleting a Node from the Linked List (assuming values in Data are unique)
Function Delete(Value)
Pointer : Found 'Pointer for finding required Node to Delete
Found = NULL 'Assuming required Node not found
IF there are no Nodes in the List then
WRITE error message that can't delete a Node
ELSE
Go to memory location pointed by Start pointer 'Begin at the Start (first Node)
Start Loop and run until Next points to NULL 'Check each Node until Next pointer points to NULL
IF value in Data of current Node is equal to Value then 'If we find our required Node
Found = current Node 'Mark the current Node with Found pointer
Exit Loop 'No need to finish the loop
ENDIF
Go to Node pointed by Next pointer of current Node
End Loop
IF the required Node is the first Node in the List then
Point the Start pointer to NULL 'Node deleted
ELSE
Point Next of previous Node to the Node pointed by Next of Found 'Node deleted (Previous.Next = Found.Next)
ENDIF
ENDIF
End Function
linear search
pseudocode for sequential search
Begin
Accept the N elements
READ Element X to search
count=0
Label1 Repeat for index=0 to N-1
IF(Element[index]=X)
WRITE Element X found
ELSE
count++
Goto Label1:
ENDIF
END LOOP
IF(count>1)
WRITE Element X not found
ENDIF
End
binary search
Begin
ACCEPT N Elements
READ Element X to search
IF (range contains only one element):
Look for desired value
IF found
WRITE Element X found
Else:
1. Get midpoint of range(N)
2. Determine which half of range contains
the value
3. Search that half of the range using binary
search
Look for desired value
IF found
WRITE Element X found
ENDIF
ENDIF
End
INSERTION SORT
READ numberofelements, array [];
FOR i=0 to numberofelements-1
J=i
WHILE j>0 and array[j]<array[j-1]
SWAP array[j] and array[j-1]
j—
WRITE array []
#include
int main()
{
int a[10],i,n,m,c=0,l,u,mid;
printf("Enter the size of an array: ");
scanf("%d",&n);
printf("Enter the elements in ascending order: ");
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
printf("Enter the number to be search: ");
scanf("%d",&m);
l=0,u=n-1;
while(l<=u)
{
mid=(l+u)/2;
if(m==a[mid])
{
c=1;
break; }
else if(m<a[mid])
{
u=mid-1; }
else
l=mid+1;
}
if(c==0)
printf("The number is not found.");
else
printf("The number is found.");
getch ();
return 0;
}
/*proram to calculate the pay of a worker*/
#include
#include
float pays(int hours,float rate);
int main()
{
int hours,itrvar,num;
float rate;
float pay=0,totalpay=0;
printf("enter the number of workers");
scanf("%d",&num);
for(itrvar=0;itrvar0)/*condition to check hourly rate should be greater than 0*/
{
pay=pays(hours,rate);/*call of pays function*/
printf("Pay at %f rupees/hr for %d is %f\n",rate,hours,pay);
totalpay=totalpay+pay;
}/*end of if*/
else
{
printf("Total pay is=%f",totalpay);
exit(0);
}/*end of else*/
}/*end of loop*/
printf("Total pay is=%f",totalpay);
}/*end of main()*/
float pays(int hours,float rate)
{
float pay;
if(hours<40)/*check for number of hours worked for calculating pay*/
pay=hours*rate;
else if(hours>=40&&hours<60)
pay=hours*1.5*rate;
else if(hours>=60)
pay=hours*2.0*rate;
return pay;
}/*end of pays()*/
/*proram to calculate the pay of a worker*/
#include
#include
float pays(int hours,float rate);
int main()
{
int hours,itrvar,num;
float rate;
float pay=0,totalpay=0;
printf("enter the number of workers");
scanf("%d",&num);
for(itrvar=0;itrvar0)/*condition to check hourly rate should be greater than 0*/
{
pay=pays(hours,rate);/*call of pays function*/
printf("Pay at %f rupees/hr for %d is %f\n",rate,hours,pay);
totalpay=totalpay+pay;
}/*end of if*/
else
{
printf("Total pay is=%f",totalpay);
exit(0);
}/*end of else*/
}/*end of loop*/
printf("Total pay is=%f",totalpay);
}/*end of main()*/
float pays(int hours,float rate)
{
float pay;
if(hours<40)/*check for number of hours worked for calculating pay*/
pay=hours*rate;
else if(hours>=40&&hours<60)
pay=hours*1.5*rate;
else if(hours>=60)
pay=hours*2.0*rate;
return pay;
}/*end of pays()*/
file pointer
#include
#include
#include
int main()
{
FILE *fp1, *fp2;
char file1[10], file2[10], data[100];
printf("\n Enter name of source file: ");
scanf("%s",&file1);
printf("\n Enter name of destination file: ");
scanf("%s",&file2);
if((fp1 = fopen(file1,"r"))==NULL)
{
printf("\n %s not found\n\n", file1);
getch();
exit(0);
}
fscanf(fp1, "%s", data);
if((fp2 = fopen(file2, "w"))==NULL)
{
printf("\n %s not found\n\n", file2);
getch();
exit(0);
}
fprintf(fp2, "%s", data);
getch();
return 0;
}
#include
#include
int main()
{
FILE *p,*q;
char b[10];
char a;
printf("enter the file name.txt");
scanf("%s",&b);
p=fopen(b,"r");
if (p==NULL)
{
printf("\n cannot locate the file");
exit ();
}
q=fopen("muddesir.txt","w");
if(q==NULL)
{
printf("\n cannot locate the file");
fclose(q);
exit ();
}
while(1)
{
a=fgetc(p);
if(a==EOF)
break;
else
fputc(a,q);
}
fclose(p);
fclose(q);
getch ();
}
#include
#include
int Factorial(int Num)
{
int loop;
for(loop=Num; loop>0; loop--)
Num*=loop;
return Num;
}
int main()
{
int n, r, Choice=0;
float Result=0.0f;
printf("\n Enter n and r values: ");
scanf("%d%d",&n, &r);
printf("\n Enter:\n\n 1 for nCr\n\n 2 for nPr\n\n Your choice: ");
scanf("%d", &Choice);
if(Choice == 1)
{
printf("\n nCr = ");
Result = (Factorial(n)/(Factorial(r) * (Factorial(n-r))));
}
else if(Choice == 2)
{
printf("\n nPr = ");
Result = Factorial(n)/Factorial(n-r);
}
else
{
printf("\n Illegal choice!");
exit(0);
}
printf("%.2f",Result);
getch();
return 0;
}
Module for AcceptValues:
1. FOR loop1=1 -> n
1.1 FOR loop2=1 -> 7
1.1.1 READ Data[loop1][loop2]
2. State[25] = {all names of states}
3. Sex[2] = {‘M’,’F’}
Module for Census:
1. FOR loop1=1 -> n
1.1 IF Data[4][loop1] = State[index1]
1.2 Print State[index1]
1.3 FOR loop2= 1 -> n
1.3.1 Index2 = 0
1.3.2 IF Data[1][loop2] = Sex[index2]
1.3.3 Print Sex[loop2]
1.3.4 FOR loop3= 1 -> n
1.3.4.1 SWITCH Data[7][loop3]
1.3.4.1.1 CASE 1 : Hindu++
1.3.4.1.2 CASE 2: Muslim++
1.3.4.1.3 CASE 3: Christian++
1.3.4.1.4 CASE 9: Others++
1.3.5 Print Hindu, Muslim, Christian, Others
1.3.6 index2++
1.4 index1++
Main() function:
1. Begin
2. GOTO AcceptValues
3. GOTO Census
4. End program
/****************
Student database
*****************/
#include
#include
/* Sorting data using bubble sort */
int* Sort(int* Marks, int Num)
{
int temp, loop1, loop2;
for(loop1=0; loop1 Marks[loop2+1])
{
temp=Marks[loop2];
Marks[loop2]=Marks[loop2+1];
Marks[loop2+1]=temp;
}
return Marks;
}
/* Searching the details based on name */
int SearchByName(char* Name, char* Value, int n)
{
int loop;
for(loop=0; loop4)
printf("\n Illegal choice!!");
else
{
switch(Choice)
{
case 1: printf("\n Enter name to search: "); /* Searching by name */
scanf("%s",Value);
temp1= SearchByName(Name, Value, Num);
if(temp1!=-1)
printf("\n Name: %s\n ID: %d\n Marks: %d\n\n", Name[temp1], ID[temp1], Marks[temp1]);
else
printf("\n Name not found");
break;
case 2: printf("\n Enter ID to search: "); /* Searching by ID */
scanf("%d", &temp2);
temp1= Search(ID, temp2, Num);
if(temp1!=-1)
printf("\n Name: %s\n ID: %d\n Marks: %d\n\n", Name[temp1], ID[temp1], Marks[temp1]);
else
printf("\n ID not found");
break;
case 3: printf("\n Students record with rank\n\n"); /* Printind data rankwise */
for(loop=0; loop<Num; loop++)
printf("\n Rank: %d\n Name: %s\n ID: %d\n Marks: %d\n\n", loop+1, Name[loop], ID[loop], Marks[loop]);
break;
case 4: printf("\n Enter ID to search: "); /* Searching name by ID */
scanf("%d", &temp2);
temp1= Search(ID, temp2, Num);
if(temp1!=-1)
printf("\n Name: %s\n\n",Name[temp1]);
else
printf("\n ID not found");
break;
}
}
getch();
return 0;
}
I tried code 2 using for loop instead of while loop, but it is not working. What could be the possible reason. I replaced i=1;
sum = 0;
while(i<n){
if(n%i==0)
sum=sum+i;
i++;
}
by
for(i=1;i<n;++i)
{
if(n%i==0)
sum+=i;
}
The compiler shows no errors but blank space after the line "The perfect numbers are:". Please reply.
I forgot to mention I also intialised sum to 0 in the beginning.
#include
#include
void mail ()
{
int n,i,sum;
clrscr();
printf("perfect number\n");
for(n=1;n<=100;n++)
{
i=1;
sum=0;
while(i<n)
{
if(n%i==0)
sum=sum+i;
i++;
}
if(sum==n)
printf("%\n",n);
}
getch();
}
Write a C program to read through an array of any type using pointers. Write a C program to scan through this array to find a particular value.
these are the question in c asked by the interviewer's so friends utilize this site.
this is a great job
#include
int is_perfect(int);
int main()
{
int x;
printf("Enter any number\n");
scanf("%d",&x);
if(is_perfect(x))
printf("perfect\n");
else
printf("not perfect");
}
int is_perfect(int x)
{
int i=1;
int sum=0;
while(i<x)
{
if(x%i==0)
sum=sum+i;
}
if(sum==0)
return 1;
else
return 0;
}
Create a program to calculate the salary and bonus based on sales of a staff.
(i) In main() :
(ii) In function get_bonus(...) :
- Ask the user to enter satff id, salary and units sold.
- Call function get_bonus(...) and pass units sold and salary to calculate
the bonus amount.
- Call function get_nett_salary(...) and pass salary and bonus amount to
calculate the nett salary.
- Call function display(...) and pass staff id, salary, units sold, bonus
amount and nett salary to be displayed on screen.
- get the units sold and salary from main() to calculate the bonus amount
by refering to the following table.
- return the bonus amount to main().
UNITS SOLD BONUS
> 1000 20% of salary
501 - 1000 10% of salary
(ii) In function get_nett_salary(...) :
(iv) In function display(...)
- get the bonus amount and salary from main() to calculate the nett salary.
- return the nett salary to main()
- get staff id, salary, units sold, bonus amount and nett salary from main.
- display all information on screen.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
DATA ENTRY
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Enter staff id : 1234
Enter staff salary : RM 3500.00
Enter total units sold : 750
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SALARY SLIP
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Staff ID : 1234
Staff salary : RM 3500.00
Units sold : 750
Bonus : RM 350.00
Nett Salary : RM 3850.00
Press any key to continue
can someone help me with this question?
how to get all program at once plz
I cant understand the logic of perfect number please expain it
please provide me pdf of all c program
in n=4 why didnt i increased to 4 after i++
in n=4 why didnt i increased to 4 after i++
When we type 0 ...it says 0 is a perfect number ...but actually 0 is not a perfect number :/ how to fix it ??
Hello Code 1 to find perfect number does not work. It gives different output. It seems like a garbage value.I tried it on different compilers too.
//Addition of two numbers without using Addition operator//
#include
#include
void main()
{
int a,b;
printf("Enter 2 no");
scanf("%d%d",&a,&b);
while(b)
{
a++;
b--;
}
printf("sum %d",a);
}
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//Addition of two numbers without using Addition operator//
#include
#include
void main()
{
int a,b;
printf("Enter 2 no");
scanf("%d%d",&a,&b);
while(b)
{
a++;
b--;
}
printf("sum %d",a);
}
www.pcprogramblog.wordpress.com
Write a program to find out first n perfect number where n is the input from user
What is perfect specifier for adding of two numbers???
Share market with certificate
Share market with certificate
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