For 10 possible values of N, values of a, b, and c will be as under

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1 CS101 Mid-semester examination Autumn Questions with sample answers Q. 1. The following program is executed with input formed by the last two digits of your Roll number, (if Roll no. is 11D50038, input is 38). Write the exact output. (4) int main(){ int R, N, k, m, a, b, c, A[10]; cin >> R; N = (R/10 + R%10) % 10; If (N == 0) { N = 10; cout << Value of N is: << N << endl; a = 0; for (k =1; k<=n; a+=k, k++); cout << a is: << a << endl; b = 1; m = 1; while (m < N/2){ b = b * m; m++; cout << b is: << b << endl; m = 1; k =0; do { if (m%n == 0){ A[k] = m; k ++; m ++; while (k <10); c = A[2]; cout << c is: << c << endl; Ans. 1. First find N, by adding the last two digits of the roll number, and then calculating modulo 10 remainder of that sum. If N = 0, reset it to 10. For 10 possible values of N, values of a, b, and c will be as under N a b c

2 Q. 2. A polynomial in x, of degree N, has been represented by storing its N +1 coefficients in an array A[50]. Its value for any x is given by A[N] + A[N-1] x + A[N-2] x 2 + A[N-3] x A[0] x N This polynomial needs to be evaluated for different values of x at different points in a program. Write a C++ function to calculate its value for any given x. (3) Ans. 2. This problem is similar to the one discussed in the class on evaluation of polynomial functions whose coefficients are stored in an array. The problem considered then was to evaluate: A[0] + A[1] x + A[2] x 2 + A[3] x A[N] x N The difference is the oder in which these coefficients appear in this question. The C++ function can be written as /* Q2 Midsem 2011 */ float poly(float A[50], int N, float x){ float sum, term;int k; sum = 0.0, term =1.0; for (k = N; k >=0; k--){ sum = sum + A[k] * term; term = term*x; return sum; 2

3 Q. 3. A value of x is given, such that 0 < x < 1. We need to find the sum of terms of the following series, such that all terms greater than 1.0E-12, are included in the sum. Write a program to read x, verify that it is within given limits, and find the desired sum. The program should also print the number of terms added in the sum. (3) sum = 1 + x + x*x + x*x*x + x* x*x*x +.. Ans. 3. /* Q3 Midsem 2011*/ #include <iostream> #include <cstring> using namespace std; int main() { float x, sum, term; int N; cout << "Give value of x" << endl; cin >> x; if ((x <= 0.0) (x >= 1.0)) { cout << "Invalid value: " << x << endl; return 1; N=0; term = 1.0; sum = 0.0; while (term > 1.0E-12){ sum = sum + term; term = term * x; N++; cout << "Sum is: " << sum << endl; cout << "No of terms added: " << N << endl; return 0; 3

4 Q. 4. An n-gram is a sub-sequence of n successive items from a given sequence, say, all characters in a word. For example, following n-grams can be formed from the word host : 1-gram: h, o, s, t (4 words of 1 character each) 2-gram: ho, os, st (3 words of 2 characters each) 3-gram: hos, ost (2 words of 3 characters each) 4-gram: host (1 word of 4 characters This is the original word) Write a program to print all the n-grams found for a given word with N characters. Compare the count of all n-grams with (N*N +N)/2, which is the actual theoretical value. (6) Ans. 4. /* Q4 Midsem 2011 */ #include<iostream> #include<cstring> using namespace std; int main(){ int k,m, j, wordlength, count=0; char word[80], ngram[50][80]; cout<<"enter a word : "; cin >> word; wordlength=strlen(word); /************************************** finding, counting and printing n-grams **************************************/ count = 0; for (k = 1; k <= wordlength; k++){ /* find all k-grams */ cout << endl << k <<"-grams: "; for (m = 0; m < wordlength -k+1; m++){ for (j =0; j < k; j++){ cout << word[m+j]; cout << ", "; count ++; cout << endl << "count is : " << count << endl; if (count == (wordlength*wordlength +wordlength)/2){ cout <<"tallies with theoretical value" << endl; else cout << "incorrect count" << endl; return 0; 4

5 Q. 5. Two arrays A[100] and B[100], contain M and N integer numbers respectively. Each array has been sorted in ascending order. Write a program to merge the contents of these two arrays into another integer array C[200], such that this resulting array is sorted in descending order. For example, if A contains 5, 7, 15; and B contains 3, 4, 13, 14, 17; then the resulting merged array C should contain 17, 15, 14, 13, 7, 5, 4, 3. Note: All legal values of integer type could be present in the array. As such, any artificial sentinel value cannot be inserted in the arrays, for use in your merge algorithm. (6) Ans. 5. This is another problem similar to one discussed in the class. There are two differences. The first is the stipulation that you cannot use any artificial sentinel value. The second is that the resultant array C[] needs to be sorted in the opposite (descending) order as compared to the order in which arrays A[] and B[] are sorted. To address the first issue, we need to tweak the merge algorithm a bit. When we reach the end of any one array, one approach is to directly copy the remaining elements of the other array into the result C[]. The second issue can be addressed in one of the two ways. The first is to start the merging process backwards. The other is to perform a regular merge in C[], getting it in ascending order, and then sort it again in descending order. The second approach, apart from requiring extra programming work, results in a much costlier operation, as it involves the sorting of the entire result. We must remember the basic objective behind a merge operation. To recapitulate, if we have to sort a large array containing N elements, the algorithm we have studied so far requires order of N 2 operations. Since the merge operation requires only an order of N operations, we want to divide the original array in two parts, each requiring only an order of (N/2) 2 operations, followed by a merge. We thus get a considerable saving in overall execution time of the entire operation. The question should be understood in this context, namely, that the two arrays A[] and B[] probably came out of a single larger array as a result of such subdivision. Re-sorting the merged array C[], essentially puts us back to square one, in that, it amounts to sorting the entire original larger array. In fact it is worse, since we would spend additional time in sorting the two sub-arrays A[] and B[]. The sample program given here merges the two arrays in reverse direction. A special mention must be made of the shorter programs which can be written using the power of the post increment/decrement operators of C++. Such usage often reduces multiple statements into a single one, thus avoiding the necessity to use pair of curly brackets to put these multiple statements together withing a 'while' or 'if' statement. For example, the two constructs given below are identical in their semantics. Version (1): if (A[i] < B[j]){ C[k]=B[j]; k = k+1; j = j-1; else { C[k]=A[i]; k = k+1; i =i-1; Version 2 if (A[i] < B[j])C[k++]=B[j--]; else C[k++]=A[i--]; 5

6 /* Q5 Midsem 2011 */ /* Based on a program written by Firuza Aibara Values that are accepted in ascending order, are stored in arrays A B. The resultant array "C" Is formed using merge algorithm, which is performed in reverse order as desired. */ #include<iostream> using namespace std; int main(){ int A[100], B[100], C[200], M, N, total; int i, j, k; // Read arrays A[] and B[] cout<<"\nenter number of elements in array A : "; cin>>m; cout<<"enter number of elements in array B : "; cin>>n; total=m+n; cout<<"\nenter "<<M<<" values for Array A in ascending order \n"; for(i=0;i<m;i++)cin>>a[i]; cout<<"\nenter "<<N<<" values for Array B in ascending order \n"; for(i=0;i<n;i++)cin>>b[i]; cout<<"\nvalues in Array A :-\n"; for(i=0;i<m;i++) cout<<a[i]<<" "; cout << "\n"; cout<<"\nvalues in Array B :-\n"; for(i=0;i<n;i++)cout<<b[i]<<" "; cout<<"\n"; // Merging array A and B in Descending order i=m-1; j=n-1; k =0; while (k < total){ if (A[i] < B[j])C[k++]=B[j--]; else C[k++]=A[i--]; if (i == 0){ // reached end of A, copy B to C while (j > 0) C[k++] = B[j--]; if (j == 0){ // reached end of B, copy A to C while (i > 0) C[k++] = A[i--]; cout<<"resultant Array C in Descending order :-\n"; for(i=0;i<total;i++)cout<<c[i]<<" "; cout<<"\n"; return 0; 6

7 Problem (for questions 6, 7, and 8). A branch of a bank has N customers (N < 50000). A customer maintains one or more accounts in that branch. Each customer has a customer-id which is a six digit unique customer number starting with 1. A 7-digit account-number is allotted for each account held by a customer. Every account number starts with 411 for that branch. Balance amount in each account is known. It is required to find out who holds maximum amount in that branch. For finding this maximum amount held by any customer, the sum of balances in all the accounts of each customer is to be considered. The data of all customers is given such that there are three entries in one line, separated by one or more blanks, containing the customer-id, account-number, and balance; and there are as many lines as there are accounts in that branch. An extra line is typed as input in the end, containing 0 for all the three values. Some sample entries are given below: (Another account for same customer , as in line 3 ) (last line containing artificial data) Write a program to successively answer the following questions. In your program, clearly mark the beginning of the portion corresponding to each question, by a comment line. Q. 6. Define 3 arrays custid[], acctno[], and balance[], to hold data for all possible accounts in that branch. Read input from each line appropriately into these arrays. There is no need to use any files, as the normal cin statement reading three values at a time, is adequate. Verify that the customer-id and account-number are valid as per prescribed rules. For any invalid value, terminate the program. Stop reading input, when you reach the last line containing artificial data (all 0 values). Count the total number of accounts, in an integer variable Nacct. (3) Q. 7. Assume that Q6 has been correctly solved. Now, compute the total balance for each customer, by adding the balance amounts in all the accounts held by that customer. Also count the number of accounts held by each customer. Define three new arrays ucustid[], totbalance[], and numacct[]. For each customer, create only one entry in each of these three arrays. For example, the entries in these arrays, for customers and will be: ucustid totbalance numacct (This customer has only one account) (The customer has two accounts, the total balance calculated) Hint: Consider sorting the original arrays so that the entries with same customer-id come as consecutive elements of array custid[]. (5) Q.8. Assume that Q7 has been correctly answered. Now, find and print the maximum balance amongst all the totbalance[] entries. Then, for each customer having this maximum balance, print the cutomer-id, account-number, and balance, for every account number belonging to that customer. (5) Ans. 6. and 7. and 8. Q. 6 is straight forward. It simply involves reading the data, and 7

8 validating that the customer-ids and the account-numbers start with '1'. and '411' respectively. The problem requires you to terminate the program if there is any invalid record. This would never be done in real life. Otherwise, you would correct the invalid record and repeat the execution, only to find another invalid input somewhere later. This could be very time consuming if you indeed have about accounts in the branch. In actual practice, the input data would always be entered into a file. The validation program would be run on this file to find all erroneous records, which are put in a separate file. The correct records are normally processed, or are stored in another file. The algorithm for question 7 is partially explained in the question itself, through the hint to sort all input records on customer id. Records for the same customer will now be in consecutive array locations. By scanning the array, we can easily assemble total balance for each customer by adding balances from consecutive records for that customer. Question 8 requires you to first fine the maximum balance of the entire lot, by using the new arrays. Now you can scan the new ucustid[] array for a customer whose balance matches the maximum, find the unique customer-id, and then search for it in the original account arrays, printing each one as you find it. The following program does the composite job for all the three questions. Some useful comments have been added for better understaning. /* Q6,7,8 midsem 2011*/ #include <iostream> using namespace std; int main() { /* Q */ int custid[50000], acctno[50000]; float balance[50000]; int k, Nacct; k = 0; Nacct =0; cin >> custid[k] >> acctno[k] >> balance[k]; cout << custid[k] << acctno[k] << balance[k]; cout << endl; while (custid[k]!=0 ){ /* received a proper entry, validate data */ if ((custid[k]/100000!= 1) (acctno[k]/10000!=411)){ cout << "invalid data" << endl; cout <<custid[k]<< " "<<acctno[k]<<" " << balance[k]; cout << endl; return 1; cout<<custid[k]<<" "<< acctno[k]<< " "<<balance[k]<<endl; Nacct++; k++; cin >> custid[k] >> acctno[k] >> balance[k]; cout <<"Total account entries are: " << Nacct << endl; 8

9 /* Q */ /* first sort data on custid */ int tempid, tempac; float tempbal; int mincustid, pos, i, j; for (i= 0; i < Nacct; i++){ mincustid = custid[i]; // assume ith element to be smallest pos =i; for (j = i+1; j <Nacct; j++){ if (custid[j] < mincustid){ pos = j; mincustid = custid[j]; tempid =custid[pos]; custid[pos]=custid[i]; custid[i]=tempid; tempac =acctno[pos]; acctno[pos]=acctno[i]; acctno[i]=tempac; tempbal=balance[pos];balance[pos]=balance[i];balance[i]=tempbal; cout << i<< " "<< custid[i]<<" "<<acctno[i]<<" "<<balance[i]; cout<< endl; /* now calculate total balance for each unique customer*/ int ucustid[50000], numacct[50000]; float totbalance[50000]; int numucust; // number of unique customers k = 0; // index for the new arrays ucustid[0] = custid[0]; numacct[0] = 1; totbalance[0] = balance[0]; // 0th entry captured as new customer for (i =1; i < Nacct; i++){ // scan all accounts if (custid[i]!= ucustid[k]){ // found new unique customer k++; ucustid[k] = custid[i]; numacct[k] = 1; totbalance[k] = balance[i]; else { numacct[k]++; totbalance[k] += balance[i]; // same customer, continue to accumulate balance numucust = k+1; // Total number of unique customers cout << endl << "Total balances" << endl; for(i=0; i<numucust; i++){ cout << i <<" " <<ucustid[i]<<" "<<numacct[i]<<" "; cout << totbalance[i] << endl; 9

10 /* Q */ /* First find the maximum balance */ float maxbalance; maxbalance = totbalance[0]; for (i=0; i < numucust; i++){ if(maxbalance < totbalance[i]) maxbalance = totbalance[i]; cout << " Maximum balance is: "<< maxbalance << endl<<endl<<endl; /* Now scan the uniqu customers for this max balance */ k=0; cout << " Customers with Maximum Balance:" << endl<<endl; for(i=0; i<numucust; i++){ if (totbalance[i]!= maxbalance) continue; k++; // else, we have found a new customer with maxbalance cout << k << ". " << "customer "<<ucustid[i]; cout << " has following account(s)" << endl; int sno = 0; for (j =0; j<nacct; j++){ //scan the accounts array if(ucustid[i] == custid[j]){ // found an account sno++; cout << " "<<sno <<" "<< acctno[j] << " "; cout << balance[j] << endl; return 0; Sample input file 'accountdata.txt'

11 Q. 9. A monochrome image of size H (Height) by W (Width) was captured in poor light. As a result, all the pixel values are known to be 130 or less. Due to some problem in the camera, rectangular white patches have got superimposed on the picture. The values of all pixels in such patches are 255. It is guaranteed that every patch is exactly rectangular. Write a program to read values of H, and W; and then values of individual pixels in an array pic[300][500]. Then find out the total number of pixels in the largest white patch (patch with largest area) present in the image. Print the start-row, start-column, and end- row, end-column of this patch. If there are multiple patches with the same size, you may print these values for any one of these patches. Two sample patches are shown in the following sketch. (5) Ans. 9. You will recall the discussions in the class on the way digital images are handled. In particular, we had seen a program which performs histogram equalization on a monochrome image. That program essentially dealt with computations involving the statistical properties of the pixel values in the image. This problem deals with the geometric placement of the pixels in a twodimensional lay-out. This question is a relatively simple version of problems of image processing involving edge detection. Since the white patches are defined to be 'rectangular', you just have to scan rows to find if any consecutive white pixels exist, and if these do exist, to find in how many subsequent columns these patches continue further. You can track the row number; and the start and end column number of each patch as you encounter it first. Additionally, keep counting the total number of pixels. Start with maxsize of 0, and maintain separately, the details of the patch with current max size. This and the Q10 will be discussed briefly in a lecture later. A problem often faced is to input large values of pixels while testing such programs. We will later see how such data can be read from a real image. But we do need large test-data with predictable behaviour to test many of our algorithms. What is included here is a program which generates artificial data for a monochrome image in a file 'picdata.txt'. It collects the coordinates of desired number of white patches from you, and replaces the corresponding pixels with value 255. Study this program to revise your understanding of simple file operations. We will be discussing more on file processing next week. 11

12 /* Creating artificial data for a monochrome image to support testing of solutions to Q9 */ #include <iostream> using namespace std; int main() { int pic[500][300]; int H, W, i,j, k, N, m, n; /* Fill up picture with artificial data*/ H =20; W = 20; for(i=0;i<h;i++) for(j=0;j<w;j++)pic[i][j]=i; /* insert white patches */ int xl[10],yl[10],xr[10],yr[10],npatches; //x, y coordiantes of uperleft and lower right corners of pathces cout << "Give number of patches"<<endl; cin >> npatches; for (i =0; i<npatches;i++){ cout << "Give x,y cordinates of top left, bottom right corner"; cin >> xl[i] >> yl[i] >> xr[i] >> yr[i]; for(k=xl[i]; k<=xr[i]; k++){ for(m=yl[i]; m<= yr[i]; m++){ pic[k][m] = 255; cout << endl; FILE* fp; fp = fopen("picdata.txt", "w"); fprintf(fp, "%d %d \n", H, W); for(i=0;i<h;i++){ for(j=0;j<w;j++){ fprintf(fp, "%d ", pic[i][j]); fprintf(fp, "\n"); fclose (fp); /* picture Data file created */ return 0; 12