Embedded Files
Some Difficult Problems
1. A program to evaluate the equation y=xn when n is a non-negative integer.
#include<stdio.h>
#include<conio.h> void main()
{
int count, n; float x,y;
printf(“Enter the values of x and n:”); scanf(“%f%d”, &x,&n);
y=1.0;
count=1; while(count<=n)
{
y=y*x; count++;
}
printf(“x=%f; n=%d; x to power n=%f”, x, n,y);
}
Output:
Output:
Enter the values of x and n: 2.5 4 X=2.500000; n=4; x to power n= 39.062500
2. A program to print the multiplication table from 1*1 to 12*10.
#include<stdio.h>
#include<conio.h>
52
#define COLMAX 10
#define ROWMAX 12 void main()
{
int row, column, y; row=1;
printf(“MULTIPLICATION TABLE ”); printf(“----------------------------------------”);
do
{
column=1; do
{
y=row*column; printf(“%d”, y); column=column +1;
}
while(column<=COLMAX); printf(“\n”);
row=row+1;
}
while(row<=ROWMAX);
printf(“---------------------------------------------”)
}
3. Program uses a for loop to print the powers of 2 table for the power 0 to 20, both positive and negative.
#include<stdio.h>
#include<conio.h> void main()
{
long int p; int n; double q;
printf(“------------------------------------------------------------”);
printf(“ 2 to power n n 2 to power -n”);
printf(“------------------------------------------------------------”); p=1;
for(n=0; n<21; ++n)
{
if(n==0)
P=1;
else
p=p*2;
}
Output:
Output:
q=1.0/(double)p;
printf(“10ld 10%d %20.12lf”, p,n,q);
}
printf(“---------------------------------------------”);
--------------------------------------------------------------------------------------------
2 to power n n 2 to power -n
----------------------------------------------------------------------------------------------
4. A class of n students take an annual examination in m subjects. A program to read the marks obtained by each student in various subjects and to compute and print the total marks obtained by each of them.
#include<stdio.h>
#include<conio.h>
#define FIRST 360
# define SECOND 240 void main()
{
int n, m, i, j, roll number, marks, total; printf(“Enter number of students and subjects”); scanf(“%d%d”, &n,&m);
printf(“\n”); for(i=1; i<=n; ++i)
{
printf(“Enter roll number:”);
scanf(“%d”, &roll_number); total=0;
printf(“Enter marks of %d subjects for ROLL NO”, m, roll number); for(j=1; j<=m; j++)
{
scanf(“%d”, &marks); total=total+marks;
}
printf(“TOTAL MARKS =%d”, total); if(total>=FIRST)
printf(“(First division)”); else if (total>=SECOND) printf(“(Second division)”); else printf(“(***FAIL***)”);
}
}
Output:
Output:
Enter number of students and subjects 3 6
Enter roll_number: 8701
Enter marks of 6 subjects for ROLL NO 8701 81 75 83 45 61 59
TOTAL MARKS =404 (First division) Enter roll_number:8702
Enter marks of 6 subjects for ROLL NO 8702 51 49 55 47 65 41
TOTAL MARKS =308(Second division) Enter roll_number: 8704
40 19 31 47 39 25
TOTAL MARKS=201(*** FAIL ***)
5. The program illustrates the use of the break statement in a C program.
#include<stdio.h>
#include<conio.h>
int m;
float x, sum, average;
printf(“This program computes the average of a set of computers”); printf(“Enter values one after another”);
printf(“Enter a NEGATIVE number at the end”); sum=0;
for(m=1;m<=1000; ++m)
{
scanf(“%f”,&x); if(x<0)
break; sum+=x;
}
average=sum/(float)(m-1); printf(“\n”);
printf(“Number of values =%d”,m-1); printf(“sum=%f”, sum); printf(“Average=%f”, average);
}
Output:
Output:
This program computes the average of a set of numbers Enter values one after another
Enter a NEGATIVE number at the end 21 23 24 22 26 22 -1
Number of values =6 Sum= 138.000000
Average=23.000000
6. A program to evaluate the series 1/1-x= 1+x+x2 +x3 +.....+xn.
#include<stdio.h>
#include<conio.h>
#define LOOP 100
#define ACCURACY 0.0001
void main()
{
int n;
float x, term, sum; printf(“input value of x :”); scanf(“%f”, &x);
sum=0;
for(term=1, n=1; n<=LOOP; ++n)
{
sum+=term; if(term<=ACCURACY) goto output;
term*=x;
}
printf(“FINAL VALUE OF N IS NOT SUFFICIENT TO ACHIEVE DESIRED ACCURACY”);
goto end; output:
printf(“EXIT FROM LOOP”);
printf(“sum=%f; no. of terms=%d”, sum,n); end:
;
}
Output:
Output:
Input value of x: .21 EXIT FROM LOOP
Sum=1.265800; No. of terms=7 Input value of x: .75
EXIT FROM LOOP
Sum=3.999774; No. of terms=34 Input value of x:.99
FINAL VALUE OF N IS NOT SUFFICIENT TO ACHIEVE DESIRED ACCURACY
7. Program illustrates the use of continue statement.
#include<stdio.h>
#include<conio.h>
#include<math.h>
int count, negative; double number, sqroot;
printf(“enter 9999 to STOP”); count=0;
negative=0; while(count<=100)
{
printf(“enter a number:”); scanf(“%lf”, &number); if(number==9999)
break; if(number<0)
{
printf(“Number is negative ”); negative++;
continue;
}
sqroot=sqrt(number);
printf(“Number=%lf square root=%lf ”, number, sqroot); count++;
}
printf(“Number of items done =%d”, count); printf(“Negative items=%d”, negative); printf(“END OF DATA”);
}
Output:
Output:
Enter 9999 to STOP Enter a number: 25.0 Number=25.000000 Square root =5.000000 Enter a number: 40.5 Number =40.500000 Square root=6.363961 Enter a number:-9
Number is negative Enter a number: 16 Number= 16.000000 Square root=4.000000 Enter a number: -14.75 Number is negative Enter a number: 80 Number=80.000000 Square root=8.944272
Enter a number: 9999 Number of items done=4 Negative items=2
END OF DATA
8. Program to print binomial coefficient table.
#include<stdio.h>
#include<conio.h>
#define MAX 10 void main()
{
int m, x, binom; printf(“m x”);
for (m=0; m<=10; ++m)
printf(“----------------------------------------------------------”); m=0;
do
{
printf(“%2d”, m); X=0; biom=1; while(x<=m)
{
if(m= =0 || x= =0) printf(“%4d”, binom); else
binom=binom* (m-x+1)/x; printf(“%4d”, binom);
}
x=x+1;
}
printf(“\n”); M=m+1’
}
while(m<=MAX);
printf(“-------------------------------------------------------”);
}
Output:
Output:
Mx
0
1
2
3
4
5
6
7
8
9
10
0
1
0
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6
15
20
15
6
1
7
1
7
21
35
35
21
7
1
8
1
8
28
56
70
56
28
8
1
9
1
9
36
84
126
126
84
36
9
1
10
1
10
45
120
210
252
210
120
45
10
1
9. Program to draw a histogram.
#include<stdio.h>
#include<conio.h>
#define N 5 void main()
{
int value[N]; int i, j, n, x;
for(n=0; n<N; ++n)
{
printf(“enter employees in group-%d: ”, n+1); scanf(“%d”, &x);
value[n]=x; printf(“%d”, value[n]);
}
printf(“\n”);
printf(“|\n”); for(n=0; n<N; ++n)
{
for(i=1; i<=3; i++)
{
if(i= =2)
printf(“ Group-%d | “, n+1); else
printf(“|”);
for(j=1; j<=value[n]; ++j) printf(“*”);
if(i= =2)
printf(“(%d)”, value[n]); else
printf(“\n”);
}
printf(“|\n”);
}
}
Output:
Output:
Enter employees in Group -1:12 12
Enter employees in Group -2:23 23
Enter employees in Group -3:35 35
Enter employees in Group -4:20 20
Enter employees in Group -5:11 11
|
|************ Group-1 |************ (12)
|************
|
|*********************** Group-2 |***********************(23)
|***********************
|
|*********************************** Group-3 |***********************************(35)
|***********************************
|
|******************** Group-4 |********************(20)
|********************
|
|*********** Group-5 |***********(11)
|***********
|
10. Program of minimum cost problem.
#include<stdio.h>
#include<conio.h> void main()
{
float p, cost, p1, cost1; for(p=0; p<=10; p=p+0.1)
{
cost=40-8*p+p*p; if(p= =0)
{
cost1=cost; continue;
}
if(cost>=cost1) break; cost1=cost; p1=p;
}
p=(p+p1)/2.0; cost=40-8&p+p*p;
printf(“MINIMUM COST=%.2f AT p=%.1f”, cost, p);
}
Output:
Output:
MINIMUM COST = 24.00 AT p=4.0
11. Program for plotting of two functions (y1=exp(-ax); y2=exp(-ax2 /2)).
#include<stdio.h>
#include<conio.h>
#include<math.h> void main()
{
int i;
float a, x, y1, y2; a=0.4;
printf(“ Y-------> ” ); printf(“0---------------------------------------------------------------------”);
for(x=0; x<5; x=x+0.25) y1=(int) (50*exp(-a*x)+0.25);
y2=(int) (50*exp(-a*x*x/2)+0.5); if(y1= =2)
{
if(x= = 2.5)
printf(“X |”); else printf(“|”);
for(i=1; i<=y1-1; ++i) printf(“”);
printf(“#”); continue;
}
if(y1>y2)
{
if(x= =2.5) printf(“X |”); else
printf(“ |”);
for(i=1; i<=y2-1; ++i) printf(“ ”);
printf(“*”);
for(i=1; i<=(y1-y2-1); ++i) printf(“_”);
printf(“0”); continue;
} if(x==2.5)
printf(“X|”); else
printf(“ |”);
for(i=1; i<=y1-1; ++i) printf(“ ”);
printf(“0”);
for(i=1; i<=(y2-y1-1); ++i) printf(“_”);
printf(“*”);
}
}
Output:
Output:
Y---->
0--------------------------------------------------------------------------------------------------------
| #
| 0---*
| 0---*
| 0---*
| 0---*
| 0---*
| 0---*
| 0---*
| 0---*
| 0---*
| #
12. Write a program using a single –subscribed variable to evaluate the following expressions:
Total =
10
2
å X2
the values of x1, x2, ..... are read from the terminal.
i = 1
#include<stdio.h>
#include<conio.h> void main()
{
int i;
float x[10], value, total;
printf(“ENTER 10 REAL NUMBERS”);
for(i=0; i<10; i++)
{
scanf(“%f”, &value); x[i]=value;
}
total=0.0;
for(i=0; i<10; i++)
total = total + x[i]*x[i]; printf(“\n”);
for(i=0; i<10; i++)
printf(“x[%2d] = %5.2f”, i+1, x[i]); printf(“total=%.2f”, total);
}
Output:
Output:
ENTER 10 REAL NUMBERS
1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 10.10
X[1]=1.10
X[2]=2.20
X[3]=3.30
X[4]=4.40
X[5]=5.50
X[6]=6.60
X[7]=7.70
X[8]=8.80
X[9]=9.90 X[10]=10.10
Total = 446.86
13. Given below is the list of marks obtained by a class of 50 students in an annual examination. 43 65 51 27 79 11 56 61 82 09 25 36 07 49 55 63 74 81 49 37 40 49 16 75 87 91 33 24 58 78 65 56 76 67 45 54 36 63 12 21 73 49 51 19 39 49 68 93 85 59
Write a program to count the number of students belonging to each of the following groups of marks: 0-9,10-19,20 -29,----,100.
#include<stdio.h>
#include<conio.h>
#define MAXVAL 50
#define COUNTER 11 void main()
{
float value [MAXVAL]; int i, low, high;
int group[COUNTER]={0,0,0,0,0,0,0,0,0,0,0}; for(i=0; i<MAXVAL; i++)
{
scanf(“%f”, &value[i]);
++group[(int)(value[i]/10)];
}
printf(“\n”);
printf(“GROUP RANGE FREQUENCEY”); for(i=0; i<COUNTER; i++)
{
low=i*10; if(i= =10) high=100; else high=low+9;
printf(“%2d%3d to %3d%d”, i+1, low, high, group[i]);
}
}
Output:
Output:
43 65 51 27 79 11 56 61 82 09 25 36 07 49 55 63 74 81 49 37 40 49 16 75 87 91 33 24 58 78
65 56 76 67 45 54 36 63 12 21 73 49 51 19 39 49 68 93 85 59 (Input data)
GROUP
1
RANGE
0 to 9
FREQUENCY
2
2
10 to 19
4
3
20 to 29
4
4
30 to 39
5
5
40 to 49
8
6
50 to 59
8
7
60 to 69
7
8
70 to 79
6
9
80 to 89
4
10
90 to 99
2
11
100 to 100
0
14. Write a program for sorting the elements of an array in descending order.
#include<stdio.h>
#include<conio.h> void main()
{
int *arr, temp, i, j, n; clrscr();
printf(“enter the number of elements in the array”); scanf(“%d”, &n);
arr=(int*)malloc(sizeof(int)*n); for(i=0;i<n;i++)
{
for(j=i+1; j<n; j++)
{
if(arr[i]<arr[j])
{
temp=arr[i]; arr[i]=arr[j]; arr[j]=temp;
}
}
printf(“Elements of array in descending order are”); for(i=0; i<n; i++);
getch();
}
Output:
Output:
Enter the number of elements in the array: 5 Enter a number: 32
Enter a number: 43 Enter a number: 23
Enter a number: 57 Enter a number: 47
Elements of array in descending order are: 57
47
43
32
23
15. Write a program for finding the largest number in an array
#include<stdio.h>
#include<conio.h> void main()
{
int *arr, i, j, n, LARGE; clrscr();
printf(“Enter the number of elements in the array”); scanf(“%d”, &n);
arr=(int*) malloc(sizeof(int)*n); for(i=0; i<n; i++)
{
printf(“Enter a number”); scanf(“%d”, &arr[i]);
} LARGE=arr[0];
for(i=1; i<n;i++)
{
if(arr[i]>LARGE) LARGE=arr[i];
}
printf(“The largest number in the array is : %d”, LARGE); getch();
}
Output:
Output:
Enter the number of elements in the array:5 Enter a number: 32
Enter a number: 43 Enter a number: 23 Enter a number: 57 Enter a number: 47
The largest number in the array is : 57
16. Write a program for removing the duplicate element in an array.
#include<stdio.h>
#include<conio.h>
#include<stdlib.h> void main()
{
int *arr, i, j, n, x, temp; clrscr();
printf(“Enter the number of elements in the array”); scanf(“%d”, &n);
arr=(int*) malloc(sizeof(int)*n); for(i=0; i<n; i++)
{
printf(“Enter a number”); scanf(“%d”, &arr[i]);
}
for(i=0; i<n; i++)
{
for(j=i+1; j<n; j++)
{
if(arr[i]>arr[j])
{
temp=arr[i]; arr[i]=arr[j]; arr[j]=temp;
}
printf(“Elements of array after sorting”); for(i=0; i<n; i++)
printf(“%d”, arr[i]); i=0;
j=1;
while(i<n)
{
if(arr[i]==arr[j])
{
for(x=j; x<n-1; x++) arr[x]=arr[x+1];
n-;
}
else
{ i++; j++;
}
}
printf(“Elements of array after removing duplicate elements”); for(i=0; i<=n; i++)
printf(“%d”, arr[i]); getch();
}
Output:
Output:
Enter the number of elements in the array:5 Enter a number: 3
Enter a number: 3 Enter a number: 4 Enter a number: 6 Enter a number: 4
Elements of array after sorting: 3
3
4
4
6
Elements of array after removing duplicate elements: 3
4
6
17. Write a program for finding the desired kth smallest element in an array.
#include<stdio.h>
#include<conio.h> void main()
{
int *arr, i, j, n, temp, k; clrscr();
printf(“enter the number of elements in the array”); scanf(“%d”, &n);
arr=(int*)malloc(sizeof(int)*n); for(i=0;i<n;i++)
{
printf(“Enter a number”); scanf(“%d”, &arr[i]);
}
for(j=i+1; j<n; j++)
{
if(arr[i]<arr[j])
{
temp=arr[i]; arr[i]=arr[j]; arr[j]=temp;
}
printf(“Elements of array after sorting”); for(i=0; i<n; i++);
printf(“%d”, arr[i]);
printf(“Which smallest element do you want to determine”); scanf(“%d”, k);
if(k<=n)
printf(“Desired smallest element is %d”, arr[k-1]); else
printf(“Please enter a valid value for finding the particular smallest element”); getch();
}
Output:
Output:
Enter the number of elements in the array:5 Enter a number: 33
Enter a number: 32 Enter a number: 46 Enter a number: 68 Enter a number: 47
Elements of array after sorting: 32
33
46
47
68
Which smallest element do you want to determine? 3 Desired smallest element is 46
18. Program to sort a list of numbers and to determine median.
#include<stdio.h>
#include<conio.h>
#define N 10
void main()
{
int i, j, n;
float median, a[N], t;
printf(“Enter the number of items”); scanf(“%d”, &n);
printf(“Input %d values ”, n); for(i=1; i<=n; i++) scanf(“%f”, &a[i]);
for(i=1; i<=n-1; i++)
{
for(j=1; j<=n-1; j++)
{
if(a[j]<=a[j+1]) t=a[j]; a[j]=a[j+1]; a[j+1]=t;
}
else continue;
}
}
if(n%2 = =0) median=(a[n/2]+a[n/2+1])/2.0; else
median=a[n/2+1]; for(i=1; mi<=n; i++) printf(“%f”, a[i]);
printf(“median is %f”, median);
}
Output:
Output:
Enter the number of items 5
Input 5 values
1.111 2.222 3.333 4.444 5.555
5.555000 4.444000 3.333000 2.222000 1.111000
Median is 3.333000
Enter the number of items 6
Input 6 values
3 5 8 9 4 6
9.000000 8.000000 6.000000 5.000000 4.000000 3.000000
Median is 5.500000
19. Program to calculate standard deviation.
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAXSIZE 100 void main()
{
int i, n;
float value [MAXSIZE], deviation; sum=sumsqr=n=0;
printf(“Input values: input -1 to end”); for(i=1; i<MAXSIZE; i++)
{
scanf(“%f”, &value[i]); if(value[i]= =-1)
break; sum+=value[i]; n+=1;
}
mean=sum/(float)n; for(i=1; i<=n; i++)
{
deviation=value[i]-mean; sumsqr+=deviation* deviation;
}
variance=sumsqr/(float)n; stddeviation=sqrt(variance); printf(“Number of items: %d”, n); printf(“Mean: %f”,mean);
printf(“Standard deviation:%f”, stddeviation);
}
Output:
Output:
Input values: input -1 to end 65 9 27 78 12 20 33 49 -1
Number of items: 8 Mean: 36.625000
Standard deviation: 23.510303
20. Program to evaluate responses to a multiple-choice test.
#include<stdio.h>
#include<conio.h>
#define STUDENTS 3
#define ITEMS 25 void main()
{
char key[items+1], response[ITEMS+1]; int count, i, student, n, correct[ITEMS+1]; printf(“Input key to the items”);
for(i=0; i<ITEMS; i++) scanf(“%c”, &key[i]);
scanf(“%c”, &key[i]); key[i]=’\0’;
for(student=1; student<=STUDENTS; student++)
{
count=0;
printf(“\n”);
printf(“Input responses of student -%d”, student); for(i=0; i<ITEMS; i++)
scanf(“%c”, &response[i]);
scanf(“%c”, &response[i]); response[i]=’\0’;
for(i=0; i<ITEMS; i++) correct[i]=0;
for(i=0; i<ITEMS; i++) if(response[i]= = key[i])
{
count=count+1; count[i]=1;
}
printf(“\n”);
printf(“student-%d”, student);
printf(“score is %d out of %d”, count, ITEMS); printf(“Response to the items below are wrong”); n=0;
for(i=0; i<ITEMS; i++) if(correct[i]= =0)
{
printf(“%d”, i+1); n=n+1;
}
if(n= =0) printf(“NIL”);
printf(“\n”);
}
}
Output:
Output:
Input key to the items abcdabcdabcdabcdabcdabcda
Input responses of student-1 abcdabcdabcdabcdabcdabcda
student-1
score is 25 out of 25
response to the following items are wrong NIL
Input responses of student-2 abcddcbaabcdabcdddddddddd
student-2
score is 14 out of 25
Response to the following items are wrong 5 6 7 8 17 18 19 21 22 23 25
Input responses of student-3 aaaaaaaaaaaaaaaaaaaaaaaa
student-3
score is 7 out of 25
Response to the following items are wrong
2 3 4 6 b7 8 10 11 12 14 15 16 18 19 20 22 23 24
21. Program for production and sales analysis.
#include<stdio.h>
#include<conio.h> void main()
{
int M[5][6], s[5][6], c[6], Mvalue[5][6], Svalue [5][6],Mweek[5], Sweek[5], Mproduct[6], Sproduct[6], Mtotal, Stotal, i, j, number;
printf(“Enter products manufactured week wise”);
printf(“M11, M12,-, M21, M22, - etc”);
for(i=1; i<=4; i++) for(j=1; j<=5; j++) scanf(“%d”, &M[i][j]);
printf(“Enter products sold week wise”); printf(“S11, S12,-, S21,S22,- etc”);
for(i=1; i<=4; i++) for(j=1; j<=5; j++) scanf(“%d”, &S[i][j]);
printf(“enter the cost of each product”); for(j=1;j<=5; j++)
scanf(“%d”, &C[j]); for(i=1; i<=4; i++) for(j=1; j<=5; j++)
{
Mvalue[i][j]=M[i][j]*c[j];
Svalue[i][j]=S[i][j]*c[j];
}
for(i=1; i<=4; i++)
{
Mweek[i]=0; Sweek[i]=0; for(j=1; j<5; j++)
{
Mweek[i] +=Mvalue[i][j];
Sweek[i]+= Svalue[i][j];
}
}
for(j=1; j<=5; j++)
{
Mproduct[j]=0; Sproduct[j]=0; for(i=1; i<=4; i++)
{
Mproduct[i] +=Mvalue[i][j];
Sproduct[i]+= Svalue[i][j];
}
Mtotal=stotal=0; for(i=1; i<=4; i++)
{
Mtotal+=Mweek[i]; Stotal+=Sweek[i];
}
printf(“\n\n”);
printf(“Following is the list of things you can ”);
printf(“request for enter appropriate item number and press RETURN key”); printf(“1. Value matrices of production & sales”);
printf(“2. Total value of weekly production & sales”); printf(“3. Product-wise monthly value of production& sales”); printf(“4. Grand total value of production & sales”); printf(“5.Exit”);
number=0; while(1)
{
printf(“ENTER YOUR CHOICE:”);
scanf(“%d”, &number); printf(“\n”);
if(number= =5)
{
printf(“GOOD BYE”); break;
}
switch(number)
{
case 1:
printf(“VALUE MATRIX OF PRODUCTION”);
for(i=1; i<=4; i++)
{
printf(“Week (%d), i”);
for(j=1; j<=5; j++) printf(“%7d”, Mvalue); printf(“\n”);
}
printf(“VALUE MATRIX OF SALES”);
for(i=1; i<=4; i++)
{
printf(“Week(%d)”, i); for(j=1; j<=5; j++) printf(“%7d”, Svalue[i][j]); printf(“\n”);
}
break; case 2:
printf(“TOTAL WEEKLY PRODUCTION & SALES”); printf(“ PRODUCTION SALES”);
printf(“ -------------- ---- ”);
for(i=1; i<=4; i++)
{
printf(“week(%d”, i);
printf(“%7d%7d”, Mweek[i], Sweek[i]);
}
break; case 3:
printf(“PRODUCT WISE TOTAL PRODUCTION & SALES”); printf(“ PRODUCTION SALES”);
printf(“ -------------- ---- ”);
for(j=1; j<=5; j++)
{
printf(“Product(%d”,ji);
printf(“%7d%7d”, Mproduct[j], Sproduct[j]);
}
break; case 4:
printf(“GRAND TOTAL OF PRODUCTION SALES”);
printf(“Total production=%d”, Mtotal); break;
default:
printf(“Wrong choice, select gain”); break;
}
}
Printf(“Exit from the program”);
}
Output:
Output:
11
15
12
14
13
13
13
14
15
12
12
16
10
15
14
14
11
15
13
12
Enter products manufactured week wise M11, M12, ----, M21, M22,------etc
10
13
9
12
11
12
10
12
14
10
11
14
10
14
12
12
10
13
11
10
Enter products sold week wise S11, S12, ----, S21, S22,-----etc
Enter cost of each product 10 20 30 15 25
Following is the list of things you can request for enter appropriate item number and press RETURN key.
1. Value matrices of production & sales
2. Total value of weekly production & sales
3. Product-wise monthly value of production & sales
4. Grand total value of production & sales
5. Exit
Enter your choice: 1
VALUE MATRIX OF PRODUCTION
Week (1)
110
300
360
210
325
Week (2)
130
260
420
225
300
Week (3)
120
320
300
225
350
Week (4)
140
220
450
210
300
VALUE MATRIX OF SALES
Week (1)
100
260
270
180
275
Week (2)
120
200
360
210
250
Week (3)
110
280
300
210
300
Week (4)
120
200
390
165
250
Enter your choice: 2
TOTAL WEEKLY PRODUCTION & SALES PRODUCTION SALES
-------------------- ----------
Week(1)
1305
1085
Week(2)
1335
1140
Week(3)
1305
1200
Week(4)
1315
1125
Enter your choice: 3
PRODUCT WISE TOTAL PRODUCTION & SALES
PRODUCTION
SALES
-------------------
---------
Product(1)
500
450
Product(2)
1100
450
Product(3)
1530
450
Product(4)
855
450
Product(5)
1275
1075
Enter your choice: 4
GRAND TOTAL OF PRODUCTION SALES
Total production=5260
Total sales=4550
ENTER YOUR CHOICE:5 GOOD BYE
Exit from the program
22. Write a program to read a series of words from a terminal using scanf function.
#include<stdio.h>
#include<conio.h> void main()
{
char word1[40], word2[40], word3[40], word4[40]; printf(“enter text:”);
scanf(“%s%s”, word1, word2); scanf(“%s”, word3);
scanf(“%s”, word4);
printf(“\n”);
printf(“word1=%s\nword2=%s\n”, word1, word2); printf(“word3=%s\nword4=%s\n”, word3, word4”);
}
Output:
Output:
Enter text:
Seventh Street, sakthinagar, erode
Word1=seventh Word2=street Word3=sakthinagar Word4=erode
23. Program to read a line of text from terminal.
#include<stdio.h>
#include<conio.h> void main()
{
char line[81], character;
int c; c=0;
printf(“Enter text. Press<return> at end”); do
{
charcter=getchar(); line[c]=character; c++;
}
while(character!=’\n’); c=c-1;
line[c]=’\0’; printf(“\n%s\n”, line);
}
Output:
Output:
Enter text.press<return> at end Programming in c is interesting Programming in c is interesting Enter text. Press <Return> at end
24. Write a program to copy one string into another and count the number of characters copied.
#include<stdio.h>
#include<conio.h> void main()
{
char string1[80], string2[80]; int i;
printf(“Enter a string\n”); printf(“?”);
scanf(“%s, string2”); for(i=0; string2[i]!=’\0’; i++) string1[i]=string2[i]; string1[i]=’\0’;
printf(“\n”);
printf(“%s\n”, string1);
printf(“number of charcters = %d\n”, i);
}
Output:
Output:
Enter a string
? Manchester Manchester
Number of characters=10
Enter a string
?westminister
Westiminister
Number of characters=11
25. Program for printing of the alphabet set in decimal and character form.
#include<stdio.h>
#include<conio.h> void main()
{
char c; printf(“\n\n”);
for(c=65; c<=122; c=c+1)
{
if(c>90&&c<97) continue;
printf(“|%4d-%c”, c,c);
}
printf(“|n”);
}
Output:
Output:
|65-A|66-B|67-C|68-D|69-E|70-F
|71-G|72-H|73-I|74-J|75-K|76-L
|77-M|78-N|79-O|80-P|81-Q|82-R
|83-S|84-T|85-U|86-V|87-W|88-X
|89-Y|90-Z|97-A|98-B|99-C|100-d
|101-e|102-f|103-g|104-h|105-i|106-j
|107-k|108-l|109-m|110-n|111-o|112-p
|113-q|114-r|115-s|116-t|117-u|118-v
|119-w|120-x|121-y|122-z|
26. Program to concatenation of strings.
#include<stdio.h>
#include<conio.h> void main()
{
int i, j, k;
char first_name[10]={ANANDA}; char second_name[10]={MURUGAN}; char last_name[10] = {SELVARAJ}; char name[30];
for(i=0; first _name[i]!=‘\0’; i++) name[i]=first_name[i]; name[i]=’ ’;
for(j=0; second _name[j]!=‘\0’;j++) name[i+j+1]=second_name[j]; name[i+j+1]=’ ’;
for(k=0; last _name[k]!=‘\0’;k++) name[i+j+k+2]=‘\0’; printf(“\n\n”);
printf(“%s\n”, name);
}
Output:
Output:
ANANDA MURUGAN SELVARAJ
27. Program to illustration of string handling functions.
#include<stdio.h>
#include<conio.h>
#include<string.h> void main()
{
char s1[20], s2[20], s3[20];
int x, l1,l2,l3;
printf(“enter two string constants”); printf(“?”);
scanf(“%s %s”, s1,s2); x=strcmp(s1,s2); if(x!=0)
{
printf(“strings are not equal”); strcat(s1, s2);
}
else
printf(“strings are equal”); strcpy(s3,s1); l1=strlen(s1); l2=strlen(s2); l3=strlen(s3);
printf(“s1=%s length =%d characters”, s1,l1); printf(“s2=%s length =%d characters”, s2,l2); printf(“s3=%s length =%d characters”, s3,l3);
}
Output:
Output:
Enter two string constants
? ananda murugan
Strings are not equal
S1=ananda murugan length=13 characters
S2=murugan length=7 characters S3=ananda murugan length=13 characters
Enter two string constants
? anand anand Strings are equal
S1=anand length=5 characters S2=anand length=5 characters S3=anand length=5 characters
28. Write a program that would sort a list of names in alphabetical order.
#include<stdio.h>
#include<conio.h>
#define ITEMS 5
#define MAXCHAR 20 void main()
{
char string[ITEMS][MAXCHAR], dummy[MAXCHAR]; int i=0, j=0;
printf(“enter names of %d items”, ITEMS); while(i<ITEMS)
scanf(“%s”, string[i++]); for(i=1; i<ITEMS; i++)
{
for(j=1; j<=ITEMS-i; j++)
{
if(strcmp(string[j-1), string[j]>0)
{
strcpy(dummy, string[j-1]); strcpy(string[j-1], string[j]); strcpy(string[j], dummy);
}
}
}
printf(“alphabetical list”); for(i=0; i<ITEMS; i++) printf(“%s”, string[i]);
}
Output:
Output:
enter names of 5 times
Ananda
murugan renuka devi shri alphabetical list
Ananda devi murugan renuka shri
29. Programs for counting of characters, words and lines in a text.
#include<stdio.h>
#include<conio.h> void main()
{
char line[81], ctr;
int i, c, end=0, characters =0, words=0. lines=0; printf(“KEY IN THE TEXT”);
printf(“GIVE ONE SPACE AFTER EACH WORD WHEN COMPLETED, PRESS RETURN”);
while(end= =0)
{ c=0;
while((ctr=getchar()) !=’\n’) line[c++]=ctr;
line[c]=’\0’;
if(line[0] = =’\0’) break;
else
{
words++;
for(i=0; line[i]!=’\0’; i++)
if(line[i] = =’ ‘ || line[i] = = ‘\t’) words++;
}
lines=lines+1;
characters =characters + strlen(line);
}
printf(“\n”);
printf(“number of lines=%d”, lines); printf(“number of words=%d”, words); printf(“number of characters=%d”, characters);
}
Output:
Output:
KEY IN THE TEXT
GIVE ONE SPACE AFTER EACH WORD WHEN COMPLETED, PRESS RETURN
Admiration is a very short-lived passion. Admiration involves a glorious obliquity of vision.
Always we like those who admire us but we do not like those whom we admire. Fools admire, but men of sense approve
Number of lines=4 Number of words=36 Number of characters=205
30. Program to alphabetize a customer list.
#include<stdio.h>
#include<conio.h>
#define CUSTOMERS 10 void main()
{
char first_name[20] [10], second_name[20][10], surname[20][10], name[20][20], telephone[20][10], dummy[20];
int i,j;
printf(“input names and telephone numbers”); printf(“?”);
for(i=0; i<CUSTOMERS; i++)
{
scanf(“%s %s %s %s”, first_name[i], second_name[i], surname[i], telephone[i]); strcpy(name[i], surname[i]);
strcat(name[i], “ ,”); dummy[0]=first_name[i][0]; dummy[1]=’\0’; strcat(name[i], dummy); strcat(name[i], “.”);
dummy[0]=second_name[i][0]; dummy[1]=’\0’;
strcat(name[i], dummy);
}
for(i=1; i<=CUSTOMERS-1; i++) for(j=1; j<=CUSTOMERS-i; j++)
if(strcmp(name[j-1], name[j])>0)
{
strcpy(dummy, name[j-1]); strcpy(name[j-1], name[j]); strcpy(name[j], dummy);
strcpy (dummy, telephone[j-1]); strcpy(telephone[j-1], telephone[j]); strcpy(telephone[j], dummy);
}
printf(“CUSTOMERS LIST IN ALPHABETICAL ORDER”);
for(i=0; i<CUSTOMERS; i++)
printf(“%-20s\t%-10s”, name[i], telephone[i]);
}
Output:
Output:
Input names and telephone numbers
? anandamurugan 9486153102
Renukadevi 9486644542
CUSTOMERS LIST IN ALPHABETICAL ORDER
anandamurugan 9486153102
Renukadevi 9486644542
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