Data Structures for Exam

7/31/2019 Data Structures for Exam

1/9

Ex no: 1120-02-2012 MULTIPLICATION OF 2D ARRAY

CODING:#includemain(){int a[3][3],b[3][3],c[3][3],i,j,k;printf("Enter the first matrix elements:");for(i=0;i


2/9

RESULT:A program was created and implemented for multiplying two matrices.

Ex no: 1220-02-2012 FIBONACCI SERIES

CODING:#includeint fib(int a, int b, int c, int n);main(){int a=-1,b=1,c=0,n;printf(Enter a number : );scanf("%d",&n);printf("Fibonacci series is");fib(a,b,c,n);}

int fib(int a, int b, int c, int n)

{if ( n == 0 )return 0;

else{c=a+b;a=b;b=c;printf(%d\n,c);fib(a,b,c,n);}}

OUTPUT:Enter a number : 10Fibonacci series is01123

58132134


3/9

RESULT:A program was created and implemented for printing Fibonacci series numb

ers using the concept of recursive.Ex no: 1323-02-2012 IMPLEMENTATION OF 3D ARRAY

CODING:#include int a[5][5][5],i,j,k;int num,qty,pri;void input();

void display();void calc();main(){int ch;input();do{

printf("\n\t1.Details\n\t2.Purchase\n\t3.Exit\nEnter your Choice: ");scanf ("%d",&ch);switch (ch){case 1:

display();break;case 2:

calc();break;

}}while(ch!=3);

}void input(){for(i=0;i


4/9

{printf ("\n\t\t\tITEM DETAILS\n\n");printf ("\n\tItem no\t\tQuantity\t\tRate\n");for(i=0;i


5/9

3.ExitEnter your choice:2

ITEM PURCHASINGEnter the item number: 3Item is available...Enter the quantity needed: 3Price of 3 items is: Rs.294/-

1.Details2.Purchase3.ExitEnter your choice:1

ITEM DETAILSItem no Qty Rate

1 6 542 9 783 1 984 12 655 7 66

1.Details2.Purchase

3.ExitEnter your choice:3

RESULT:A program was created and implemented for print purchase details using 3

D array.Ex no: 1401-03-2012 INFIX TO POSTFIX CONVERSION

CODING:

#includemain(){

char exp[25];int br[10],br2[10];int op[10];int i,j,k,l1,l2,t1,t2,n;printf("\nEnter the expression string\n");scanf("%s",exp);for (i=0,j=0,k=0,n=0; exp[i]!='\0'; i++){if (exp[i] == '('){

br[j] = i;j++;

}else if (exp[i]=='+' || exp[i]=='-' || exp[i]=='*' || exp[i]=='/'){op[k] = i;k++;

}else if(exp[i]==')'){


6/9

br2[n]=i;l1 = br2[n];l2 = op[k-1];exp[l1] = exp[l2];l1--;for (;l1>=l2;l2++)

exp[l2] = exp[l2+1];exp[l2-1] = ')';n++;k++;

}printf("\nAfter %d iteration : %s",i+1,exp);

}printf("\nPostfix string is %s",exp);}

OUTPUT:Enter the expression string: ((2*6)+(8*5))

After 1 iteration: ((2*6)+(8*5))After 2 iteration: ((2*6)+(8*5))After 3 iteration: ((2*6)+(8*5))After 4 iteration: ((2*6)+(8*5))After 5 iteration: ((2*6)+(8*5))

After 6 iteration: ((26)*+(8*5))After 7 iteration: ((26)*+(8*5))After 8 iteration: ((26)*+(8*5))After 9 iteration: ((26)*+(8*5))After 10 iteration: ((26)*+(8*5))After 11 iteration: ((26)*+(8*5))After 12 iteration: ((26)*+(85)*)After 13 iteration: ((26)*(85)*+)

Postfix string is: ((26)*(85)*+)

RESULT:A program was implemented for convert infix express to a postfix express

ion.Ex no: 1501-03-2012 EVALUATION OF POSTFIX EXPRESSION

CODING:#includemain(){

char exp[25];

int br[10],br2[10];int op[10];int i,j,k,l1,l2,t1,t2,n;printf("\nEnter the expression string\n");scanf("%s",exp);for (i=0,j=0,k=0,n=0; exp[i]!='\0'; i++){if (exp[i] == '('){br[j] = i;


7/9

j++;}else if (exp[i]=='+' || exp[i]=='-' || exp[i]=='*' || exp[i]=='/'){op[k] = i;k++;

}else if(exp[i]==')'){br2[n]=i;l1 = br2[n];l2 = op[k-1];exp[l1] = exp[l2];l1--;for (;l1>=l2;l2++)

exp[l2] = exp[l2+1];exp[l2-1] = ')';n++;k++;

}printf("\nAfter %d iteration : %s",i+1,exp);

}printf("\nPostfix string is %s",exp);}

OUTPUT:Enter the expression string: ((2*6)+(8*5))

After 1 iteration: ((2*6)+(8*5))After 2 iteration: ((2*6)+(8*5))After 3 iteration: ((2*6)+(8*5))After 4 iteration: ((2*6)+(8*5))After 5 iteration: ((2*6)+(8*5))After 6 iteration: ((26)*+(8*5))After 7 iteration: ((26)*+(8*5))After 8 iteration: ((26)*+(8*5))After 9 iteration: ((26)*+(8*5))

After 10 iteration: ((26)*+(8*5))After 11 iteration: ((26)*+(8*5))After 12 iteration: ((26)*+(85)*)After 13 iteration: ((26)*(85)*+)

Postfix string is: ((26)*(85)*+)

Ex no: 1605-03-2012 Towers of Hanoi

CODING:#includemain(){void tower(int n,char source,char dest,char auxilary);int numdisk;printf("\n Enter the number of disks: ");


8/9

scanf("%d",&numdisk);printf("\t\t\tSTART TOWER OF HANOI");tower(numdisk,'a','b','c');

}void tower(int n,char source,char dest,char auxilary){if(n==1){

printf("\n\tMOVE FROM %c TO %c\n",source,dest);}else{

tower(n-1,source,auxilary,dest);printf("\n\tMOVE FROM %c TO %c\n",source,dest);tower(n-1,auxilary,dest,source);

}}

OUTPUT:

Enter the number of disks:2

START OF TOWER OF HANOI

MOVE FROM a TO cMOVE FROM a TO bMOVE FROM c TO b

RESULT:

Ex no: 1705-03-2012 ACKERMANN

CODING:#includeint m,n,h;main(){printf("Enter the value of M and N: ");scanf("%d%d",&m,&n);acker(m,n);display(m,n);

}acker(int m,int n){

if(m==0&&n!=0)

{h=n+1;

}else if(m!=0&&n==0){acker(m-1,1);

}else if(m!=0&&n!=0){acker(m-1,acker(m,n-1));


9/9

}}display(int m,int n){printf("A(%d,%d)=%d\n",m,n,h);

}

OUTPUT:Enter the value of M and N:1 1A(1,1)=3

RESULT:

Data Structures for Exam

Documents

Transcript of Data Structures for Exam