BITS PILANI, DUBAI CAMPUS DUBAI INTERNATIONAL ACADEMIC CITY, DUBAI FIRST SEMESTER

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1 BITS PILANI, DUBAI CAMPUS DUBAI INTERNATIONAL ACADEMIC CITY, DUBAI FIRST SEMESTER COURSE : COMPUTER PROGRAMMING (CS F111) COMPONENT : Tutorial#4 (SOLUTION) DATE : 09-NOV-2017 Answer 1(a). #include<stdio.h> int a[15], posn[5], key, size, k; scanf( %d,&size); scanf( %d,&key); for(i=0;i<=n-1;i++) scanf( %d,&a[i]); j=0; for(i=0;i<=n-1;i++) if(key==a[i]) posn[j]=i+1; j++; if(j==0) printf( \nkey NOT FOUND!\n\n ); else printf( Occurences of key are as follows:\n ); for(k=0;k<j;k++) printf( %d\n,posn[k]); Answer 1(b). In the previous program posn[--j] will be the last occurrence of the key. Answer 1(c). Index of all occurrences of key in a 2-D array int a[10][10], key, nrows, ncols, i, j; printf( Enter the rows and columns: ); scanf( %d %d, &nrows, ncols); for(i=0; i<nrows;i++) CS F111 COMPUTER PROGRAMMING Page 1 of 7

2 scanf( %d,&a[i][j]); scanf( %d,&key); for(i=0;i<nrows;i++) if(key==a[i][j]) printf( row no: %d\tcol no.:%d\n,i+1,j+1); Answer 1(d). The above program can be modified to get the last occurrence of the key in the 2-D array as follows int a[10][10], key, nrows, ncols, i, j,k,l; printf( Enter the rows and columns: ); scanf( %d %d, &nrows, ncols); for(i=0; i<nrows;i++) scanf( %d,&a[i][j]); scanf( %d,&key); for(i=0;i<nrows;i++) if(key==a[i][j]) k=i+1; l=j+1; printf( The last occurrence is at row no: %d\tcol no.:%d\n,k,l); Answer 2. int a[10][10], b[10]; printf( Enter the rows and columns: ); scanf( %d %d, &nrows, ncols); for(i=0; i<nrows;i++) scanf( %d,&a[i][j]); for(j=0; j<=ncols-1; j++) k=0; for(i=0; i<nrows-1; i++) b[k++]=a[i][j]; /*Perform insertion sort on b array k=0; CS F111 COMPUTER PROGRAMMING Page 2 of 7

3 for(i=0; i<=nrows-1; i++) a[i][j]=b[k++]; Answer 3. Hint: a. Implement Linear Search for each row, b. with maximum number of elements in each row as the limit, Once the Row is finalized, the c. Use Binary Search for that row and d. Use Reverse string program a. Program for Binary Search int c, first, last, middle, n, search, array[100]; printf("enter number of elements\n"); scanf("%d",&n); printf("enter %d integers\n", n); for (c = 0; c < n; c++) scanf("%d",&array[c]); printf("enter value to find\n"); scanf("%d", &search); first = 0; last = n - 1; middle = (first+last)/2; while (first <= last) if (array[middle] < search) first = middle + 1; else if (array[middle] == search) printf("%d found at location %d.\n", search, middle+1); break; else last = middle - 1; middle = (first + last)/2; if (first > last) printf("not found! %d is not present in the list.\n", search) return 0; b. Program for Linear Search #include<stdio.h> int a[10],i,n,m,c=0; printf("enter the size of an array: "); scanf("%d",&n); printf("enter the elements of the array: "); for(i=0;i<=n-1;i++) scanf("%d",&a[i]); printf("enter the number to be search: "); scanf("%d",&m); for(i=0;i<=n-1;i++) if(a[i]==m) c=1; break; if(c==0) printf("the number is not in the list"); else printf("the number is found"); return 0; CS F111 COMPUTER PROGRAMMING Page 3 of 7

4 Answer 4 #include <math.h> int isprime(int); int num, flag; printf ("Pl. enter the num : "); scanf("%d", &num); flag = isprime(num); /*Pass by value /* if ( flag == 1 ) printf ("%d is a prime no\n"); else printf("%d is not a prime no\n"); printf("%d is %s a prime no.\n", num, (flag)? "" : "not" ); int isprime(int num) int limit = sqrt(num), i; for (i = 2; i <= limit; i++) if ( (num % i) == 0 ) return 0; return 1; /* When compiled without -lm $ cc Ans1.c /tmp/cccm49mg.o: In function `isprime': Ans1.c:(.text+0x76): undefined reference to `sqrt' collect2: error: ld returned 1 exit status When compiled with -lm, no errors $ cc Ans1.c lm (This is linking math library for sqrt() function) Pl. enter the num : is a prime no. Pl. enter the num : is a prime no. Pl. enter the num : is not a prime no. Answer 5 #define ORIGIN 0 #define QUADRANT_1 1 #define QUADRANT_2 2 CS F111 COMPUTER PROGRAMMING Page 4 of 7

5 #define QUADRANT_3 3 #define QUADRANT_4 4 int whichquadrant(int x, int y) int quadrant; if (!x &&!y) return ORIGIN; quadrant = (x>0)? (y>0)? QUADRANT_1 : QUADRANT_4 : (y>0)? QUADRANT_2 : QUADRANT_3; return quadrant; int x, y, quadrant; printf ("Pl, enter co-ordinates x,y of a point: "); scanf ("%d %d", &x, &y); quadrant = whichquadrant(x, y); if (quadrant == ORIGIN) printf ("The point with co-ordinates x=%d & y=%d is the origin\n",x,y); else printf ("The point with co-ordinates x=%d & y=%d is in quadrant no. %d\n", x, y, quadrant); /* $ cc Ans2.c Pl, enter co-ordinates x,y of a point: 0 0 The point with co-ordinates x=0 & y=0 is the origin Pl, enter co-ordinates x,y of a point:1 1 The point with co-ordinates x=1 & y=1 is in quadrant no. 1 Pl, enter co-ordinates x,y of a point: The point with co-ordinates x=-1 & y=10 is in quadrant no. 2 Pl, enter co-ordinates x,y of a point: 4-1 The point with co-ordinates x=4 & y=-1 is in quadrant no. 4 Pl, enter co-ordinates x,y of a point: -5-6 The point with co-ordinates x=-5 & y=-6 is in quadrant no. 3 Answer 6 /* A). Pl note that variables x, y, z, min in function findmin() is different from the variables x, y, z, & min in function, even though the names are same. This is because the "scope" of variables x, y, and z in function is between the of and that of x, y, and z in function findmin() is between of findmin(). B). Also note that the variables x, y, and z are passed by value to function findmin(), So any changes in values of x, y and z made in function findmin() will be localized (i.e. will be seen only inside the function). CS F111 COMPUTER PROGRAMMING Page 5 of 7

6 int findmin(int x, int y, int z) int min = 0; printf (" In value of x=%d, y=%d, z=%d, min=%d\n", x, y, z, min = x; if (y < min ) min = y; if (z < min ) min = z; if (min == x) x--; y++; z++; else if (min == y) x++; y--; z++; else x++; y++; z--; printf (" In value of x=%d, y=%d, z=%d, min=%d\n", x, y, z, return min; int x, y, z, min = 0; printf("pl. enter 3 nos : "); scanf("%d %d %d", &x, &y, &z); printf ("Before calling findmin: value of x=%d, y=%d, z=%d, min=%d\n", x, y, z, min = findmin(x, y, z); printf ("After calling findmin: value of x=%d, y=%d, z=%d, min=%d\n", x, y, z, /* $ cc Ans3.c Pl. enter 3 nos : Before calling findmin: value of x=10, y=20, z=30, min=0 In value of x=10, y=20, z=30, min=0 In value of x=9, y=21, z=31, min=10 After calling findmin: value of x=10, y=20, z=30, min=10 Pl. enter 3 nos : Before calling findmin: value of x=20, y=10, z=30, min=0 In value of x=20, y=10, z=30, min=0 In value of x=21, y=9, z=31, min=10 After calling findmin: value of x=20, y=10, z=30, min=10 Pl. enter 3 nos : Before calling findmin: value of x=20, y=30, z=10, min=0 In value of x=20, y=30, z=10, min=0 In value of x=21, y=31, z=9, min=10 CS F111 COMPUTER PROGRAMMING Page 6 of 7

7 After calling findmin: value of x=20, y=30, z=10, min=10 Answer 7 int power(int); float sumofseries(int); int n; float sum; printf("pl. enter the value of n : "); scanf("%d", &n); sum = sumofseries(n); printf("sum of series = %f\n", sum); int power(int a) int res = 1; while (a--) res *= 2; return res; float sumofseries(int n) int i; float sum = 0; for(i=0; i<=n; i++) sum += (float)(3+2*i)/power(i); return sum; /* $ cc Ans4.c Pl. enter the value of n : 0 Sum of series = Pl. enter the value of n : Pl. enter the value of n : 2 1 Sum of series = Sum of series = ***END*** CS F111 COMPUTER PROGRAMMING Page 7 of 7