- WAP to enter the current date in dd/mm/yyyy format and the day name on the 1st of January of that year. Display the name of the day on the given date.
/* Write a program to accept a date in the string format dd/mm/yyyy * and accept the name of the day * on 1st of January of the corresponding year. Find the day for the given date. Example: INPUT Date: 5/7/2001 Day on 1st January : MONDAY OUTPUT Day on 5/7/2001 : THURSDAY Test run the program on the following inputs: INPUT DATE DAY ON 1ST JANUARY OUTPUT DAY FOR INPUT DATE 4/9/1998 THURSDAY FRIDAY 31/8/1999 FRIDAY TUESDAY 6/12/2000 SATURDAY WEDNESDAY The program should include the part for validating the inputs namely the date and the day on 1st January of that year. */ import java.util.*; class CurrentDayName { public static void main (String args[]) throws InputMismatchException { Scanner scan = new Scanner(System.in); int i,dd,mm,yy,tdays, index,r; String date,dayName; System.out.println("Enter a date in the string format dd/mm/yyyy: "); System.out.println("If the date is 9th Februaray, 2013 enter it as 09/02/2013. "); date = scan.next(); dd=mm=yy=0; // Convert the date in string format to day, month and year try{ dd = Integer.parseInt(date.substring(0,2)); mm = Integer.parseInt(date.substring(3,5)); yy = Integer.parseInt(date.substring(6)); System.out.println("Enter the name of the day on 1st of January "); dayName = scan.next(); //Store the days of the months int dom[] = {31,28,31,30,31,30,31,31,30,31,30,31}; //If it is a leap year, februrary should have 29 days. if(yy%4 == 0) dom[1] = 29; if(dd < 0 || dd > dom[mm-1]){ System.out.println("Invalid date."); }else{ //Store the day names of the week String days[] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"}; //Find the corresponding day of the year tdays = 0; //Add the days of the month for(i = 0; i < mm-1; i++) tdays+= dom[i]; //Add the days of the current month tdays+= dd; index = 0; //Find the index of the entered day name in the day names array for(i = 0; i < 7; i++) { if(dayName.equalsIgnoreCase(days[i])){ index = i; break; } } //find the index of the day in the day names array for the current day r = tdays%7 + index - 1; if( r >= 7) r -= 7; System.out.println("Day on " + date + " : " + days[r]); } // else }catch(Exception e){ System.out.println("Please enter the date in the specified format."); System.out.println("Error : "+e); } }//end of main }//end of class - WAP to enter two dates of the same year and find the number of days elapsed between them.
/* Enter two dates of the same year and find the * number of days elapsed between them */ import java.util.*; class TwoDateDiff { public static void main (String args[])throws InputMismatchException { Scanner scan = new Scanner(System.in); int d1,m1,d2,m2,y,i,total1,total2,diff; System.out.println("Enter the day and month of first date: "); d1 = scan.nextInt(); m1 = scan.nextInt(); System.out.println("Enter the day and month of second date: "); d2 = scan.nextInt(); m2 = scan.nextInt(); System.out.println("Enter the year: "); y = scan.nextInt(); int dom[] = {31,28,31,30,31,30,31,31,30,31,30,31}; //If it is a leap year, februrary should have 29 days. if(y%4 == 0) dom[1] = 29; // Check for validity of date if(d1 < 0 || m1 > 12 || d1 > dom[m1-1] || d2 < 0 || m2 >12 || d2 > dom[m2-1]){ System.out.println("Please enter a valid date."); }else{ total1 = 0; //Add the days of the month for(i=0; i< m1-1; i++) total1 += dom[i]; //Add the days of the current month total1+=d1; total2 = 0; //Add the days of the month for(i=0; i< m2-1; i++) total2 += dom[i]; //Add the days of the current month total2+=d2; diff = (int) Math.abs(total1 - total2); System.out.println("Days elapsed : "+diff); }//else }//end of main }//end of class - WAP to enter a paragraph which may be terminated by a '?', '.' or an '!'. Print the number of sentences that have 2 or more palindrome words in it.
/* * WAP to enter a paragraph which may be terminated by a '?', '.' or an '!'. * Print the number of sentences that have 2 or more palindrome words in it. */ import java.io.*; class NoOfPalindromes { public static void main (String args[])throws IOException { BufferedReader br=new BufferedReader(new InputStreamReader (System.in)); int i,p,j,f,l,count, nos=0; String str, word, revWord; char ch, ch1; System.out.print("Enter a paragraph(terminated by '?', '.' or '!') : "); str = br.readLine(); l= str.length(); p=0; for(i=0;i< l;i++) { ch=str.charAt(i); // check for the end of a sentence if(ch=='!' ||ch=='?'|| ch=='.') { f=0; count=0; //Loop to traverse through each sentence for(j=p;j< i+1;j++) { ch1=str.charAt(j); //Extract a word if(ch1==' '||ch1=='!' ||ch1=='?'|| ch1=='.') { word=str.substring(f,j); //Reverse the word revWord = (new StringBuffer(word).reverse()).toString(); //Count palindrome words if(word.equals(revWord)) count++; f=j+1; } } // If the sentence has two or more palindrome words increment // the counter if(count>=2) nos++; p=i+1; } } System.out.println("Number of sentences with two or more palindrome words in it : "+nos); }//end of main }//end of class - WAP to enter two sentences and print the common words in them.
/* * WAP to enter two senteces from the user. * Print the common words in them. */ import java.io.*; class CommonWords { public static void main (String args[])throws IOException { BufferedReader br=new BufferedReader(new InputStreamReader (System.in)); int i,j,l1,l2,p,x,y,count=0; String str1, str2; char ch; System.out.print("Enter two sentences terminated by either a '?', '.' or '!' : "); str1 = br.readLine(); str2 = br.readLine(); l1= str1.length(); l2= str2.length(); String s1[] = new String[l1]; x=0; p=0; //Store all the words of the first sentence in a string array for(i=0;i< l1;i++) { ch = str1.charAt(i); if(ch == ' ' || ch == '?' || ch == '.' || ch == '!'){ s1[x++]=str1.substring(p,i); p = i+1; } } String s2[] = new String[l1]; y=0; p=0; //Store all the words of the second sentence in a string array for(i=0;i< l2;i++) { ch = str2.charAt(i); if(ch == ' ' || ch == '?' || ch == '.' || ch == '!'){ s2[y++]=str2.substring(p,i); p = i+1; } } //Now compare the words stored in two arrays for(i=0;i< x;i++){ for(j=0;j< y;j++){ //If match is found, print the word and store blank space to avoid repetition if(s1[i].equalsIgnoreCase(s2[j]) && !s2[j].equals(" ")){ System.out.println(s1[i]); s2[j]=" "; } } } }//end of main }//end of class - Enter n sentences and merge them to form a single sentence, where n <= 10. The merged sentence should not have duplicate words and must only consist of letters and blank spaces. Any other character should be removed.
/* * Enter n sentences and merge them to form a single sentence, * where n <= 10. The merged sentence should not have duplicate words and must * only consist of letters and blank spaces. Any other character * should be removed. */ import java.io.*; class MergeSentence { public static void main (String args[])throws IOException { BufferedReader br=new BufferedReader(new InputStreamReader (System.in)); int n,i,j,k,p,x,l,now; String ms; char ch,t; System.out.println("Enter the no of sentences (upto 10): "); n = Integer.parseInt(br.readLine()); if(n<=10){ //String array to store the sentences String str[] = new String[n]; //counter to store number of words now=0; System.out.println("Enter the "+n+" sentences: "); for(i=0;i< n;i++) { str[i]=br.readLine(); //This loop is used to count the number of words in all the sentences. //The counter is then used to initialize the array that stores all the words for(j=0;j< str[i].length();j++){ ch = str[i].charAt(j); if(ch == ' ' || ch == '?' || ch == '.' || ch == '!'){ now++; } } } //String array to store number of words String words[] = new String[now]; x=0; for(i=0;i< n;i++){ l=str[i].length(); p=0; for(j=0;j< l;j++){ ch = str[i].charAt(j); ms=""; if(ch == ' ' || ch == '?' || ch == '.' || ch == '!'){ //Loop to access the word for(k=p;k< j;k++){ t = str[i].charAt(k); //Only add letters to the merged string if((t >= 'A' && t<='Z') || (t>='a' && t<='z')) ms = ms + t; } //Store the words in string array words[x++] = ms; p = j+1; } } } //Now remove duplicate words from the string array for(i=0;i< x;i++){ for(j=i+1;j< x;j++){ if(words[i].equalsIgnoreCase(words[j])){ for(k=j;k< x-1;k++) words[k] = words[k+1]; j--; // Decrement loop counter to cater to successive duplicates x--; } } } for(i=0;i< x;i++){ System.out.print(words[i]+" "); } }else{ System.out.println("Invalid value!"); } }//end of main }//end of class - Enter date of announcement, number of days given to complete the project and find the date of submission of project.
/* Enter date of announcement, number of days given to complete the project and * find the date of submission of project. */ import java.util.*; class DateOfSubmission { public static void main (String args[])throws InputMismatchException { Scanner scan = new Scanner(System.in); int dd, mm, yy, nod,i,tdays=0,m; System.out.println("Enter the day, month and year of the date of announcement of the project: "); dd = scan.nextInt(); mm = scan.nextInt(); yy = scan.nextInt(); System.out.println("Enter the no of days given to complete the project: "); nod = scan.nextInt(); int dom[] = {31,28,31,30,31,30,31,31,30,31,30,31}; //If it is a leap year, februrary should have 29 days. if(yy%4 == 0) dom[1] = 29; //Add the days of the month for(i=0; i< mm-1; i++) tdays+=dom[i]; //Add the days of the current month tdays+=dd; //Add the number of days given to complete the project tdays+=nod; //Now convert the total calculated days to date // If the days exceed the total number of days in a year // Increment the year by one if(yy%4 == 0){ if(tdays > 366){ tdays-=366; yy++; } }else{ if(tdays > 365){ tdays-=365; yy++; } } if(yy%4 == 0) dom[1] = 29; else dom[1] = 28; m=0; while(tdays>dom[m]){ tdays-=dom[m]; m++; } System.out.println("Date of submission: "+tdays+"/"+(m+1)+"/"+yy); }//end of main }//end of class - A number with n digits, which, when multiplied by 1, 2, 3, ..., n produces the same digits in a different order is known as a Cyclic Number. For example, 142857 is a cyclic number: 142857 x 2 = 285714; 142857 x 3 = 428571; 142857 x 4 = 571428; 142857 x 5 = 714285; 142857 x 6 = 857142, and so on. Write a program to enter a number and check if it is a cyclic number or not.
/* * Write a program to enter a number and check if it is a cyclic number or not. */ import java.io.*; class CyclicNo { //Function to arrange the digits stored in the string s in ascending order static String sort(String s) { String t=""; int i,j; char a[] = s.toCharArray(); for(i = 0; i < a.length; i++) { for(j = 0; j < a.length-i-1; j++) { if(a[j] > a[j+1]) { char r = a[j]; a[j] = a[j+1]; a[j+1] = r; } } } for(i = 0; i < a.length; i++) t = t + a[i]; return t; } public static void main(String arg[])throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String w,r,p; int i,n; boolean flag = true; System.out.println("Enter a number"); n = Integer.parseInt(br.readLine()); // Convert the integer to string String str = Integer.toString(n); // Sort the number w = sort(str); for(i = 2; i <= str.length(); i++) { // Get the multiples of the number r = Integer.toString(n*i); // Sort the multiple p = sort(r); if(p.equals(w) == false) { flag = false; break; } } if(flag) System.out.println(n + " is a cyclic number"); else System.out.println(n + " is not a cyclic number"); }//end of main }//end of class - A number that is smaller than the sum of its aliquot parts (proper divisors) is known as an Abundant Number. Twelve is the smallest abundant number - the sum of its aliquot parts is 1 + 2 + 3 + 4 + 6 = 16 - followed by 18, 20, 24, and 30.
/* * WAP to print all the abudant numbers between m and n. * A weird number is one which is smaller than the sum of its aliquot parts. */ import java.util.*; class AbundantNo { public static void main (String args[])throws InputMismatchException { Scanner scan = new Scanner(System.in); int m,n,i,j,s; System.out.println("Enter the limits m and n: "); m = scan.nextInt(); n = scan.nextInt(); for(j=m;j<=n;j++){ s=0; for(i=1;i< j;i++){ if(j%i == 0) s=s+i; } if(s>j) System.out.print(j+" "); } }//end of main }//end of class - A number such that both the sum of its divisors and the number of its divisors are perfect numbers is known as a Sublime Number. The smallest sublime number is 12. There are 6 divisors of 12 - 1, 2, 3, 4, 6, and 12 - the sum of which is 28. Both 6 and 28 are perfect. Write a program to check if a number is a sublime number or not.
/* * WAP to check if a number is sublime number or not. */ import java.io.*; class SublimeNo { public static void main (String args[])throws IOException { int s=0,f=0,s1=0,s2=0,n,i,j; BufferedReader obj=new BufferedReader (new InputStreamReader(System.in)); System.out.println("Enter a no: "); n=Integer.parseInt(obj.readLine()); for(i=1;i<=n;i++) { if(n%i==0) { s+=i; f++; } } for(j=1;j< s;j++) { if(s%j==0) s1+=j; } for(j=1;j< f;j++) { if(f%j==0) s2+=j; } if(s1==s && s2==f) System.out.println(n+ " is a sublime number"); else System.out.println(n+ " is not a sublime number"); } } - Find 1's complement of a number
/* * WAP to enter a negative number in base 10 and find * its 1's complement. */ import java.util.*; class OnesComplement { public static void main (String args[])throws InputMismatchException { Scanner scan = new Scanner(System.in); int n,i,p,b,x; System.out.println("Enter a number (in base 10): "); n = scan.nextInt(); if(n< 0){ int a[] = new int[100]; p = -1*n; x=0; //Convert the number to binary while(p>0){ b = p%2; a[x++] = b; p=p/2; } for(i=0; i< x; i++){ if(a[i] == 0){ a[i] = 1; }else{ a[i] = 0; } } for(i=x-1; i>=0; i--){ System.out.print(a[i]); } }else{ System.out.println("Invalid number. Please enter a negative number."); } }//end of main }//end of class - Find 2's complement of a numbeR
/* * WAP to enter a negative number in base 10 and * find its 2's complement. */ import java.util.*; class TwosComplement { public static void main (String args[])throws InputMismatchException { Scanner scan = new Scanner(System.in); int n,i,p,b,x,carry; System.out.println("Enter a number (in base 10): "); n = scan.nextInt(); if(n<0){ int a[] = new int[100]; p = -1*n; x=0; //Convert the number to binary while(p>0){ b = p%2; a[x++] = b; p=p/2; } //Convert the array to store the 1s complement for(i=0; i< x; i++){ if(a[i] == 0){ a[i] = 1; }else{ a[i] = 0; } } System.out.print("1s complement: "); for(i=x-1; i>=0; i--){ System.out.print(a[i]); } //Add 1 to the last digit of 1s complement to get 2s complement a[0] += 1; carry = 0; for(i=0; i < x; i++){ a[x] = a[x] + carry; if(a[x] == 2){ a[x] = 0; carry = 1; } } System.out.print("\n2s complement: "); for(i=x-1; i>=0; i--){ System.out.print(a[i]); } }else{ System.out.println("Invalid number. Please enter a negative number."); } }//end of main }//end of class - Add two binary numbers
/* * Enter two numbers in base 10, convert them in binary * and find their sum */ import java.io.*; class BinarySum { public static void main(String args[])throws IOException { int m,n,u,v,x,y,z,p,r,i,d,e; BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); System.out.print("Enter the first number (base 10): "); m= Integer.parseInt(br.readLine()); System.out.print("Enter the second number (base 10): "); n= Integer.parseInt(br.readLine()); //Arrays to store numbers in binary form int a[]= new int [20]; int b[]=new int [20]; //Array to store the sum of binary numbers int c[]= new int[50]; p=m; x=0; //Convert the number to binary and store in an array while(p>0) { a[x++]=p%2; p=p/2; } System.out.print(m+"="); for(i=x-1;i>=0;i--) { System.out.print(a[i]); } System.out.println(); p=n; y=0; //Convert the number to binary and store in an array while(p>0) { b[y++]=p%2; p=p/2; } System.out.print(n+"="); for(i=y-1;i>=0;i--) { System.out.print(b[i]); } e=z=u=v=0; /* * To understand the following logic, you need to know binary addition. * In binary addition: * 0 + 0 = 0 * 0 + 1 = 1 * 1 + 0 = 1 * 1 + 1 = 10 (read as 0 CARRY 1) * 1 + 1 + 1 = 11 ( read as 1 CARRY 1) */ do{ d=a[u]+b[v]+e; //Add the digits if(d==2){ c[z++]=0; e=1; } else if(d==3){ c[z++]=1; e=1; } else{ c[z++]=d; e=0; } u++; v++; }while(u < x && v < y); //If any digit is left in the first array copy them // into the sum array after adding the carry if(u < x){ while(u < x){ d=a[u++]+e; if(d==2){ c[z++]=0; e=1; } else if(d==3){ c[z++]=1; e=1; } else{ c[z++]=d; e=0; } } } //If any digit is left in the second array copy them // into the sum array after adding the carry else if(v < y){ while(v < y){ d=b[v++]+e; if(d==2){ c[z++]=0; e=1; } else if(d==3){ c[z++]=1; e=1; } else{ c[z++]=d; e=0; } } } //If a carry is still left copy it at the last in the sum array if(e==1){ c[z++]=e; } System.out.print("\nSum of the two binary numbers : "); for(i=z-1;i>=0;i--) System.out.print(c[i]); }//end of main }//end of class - Enter a sentence and print it in rotation of words for all possible combinations. Ex:
Input: java is robust Output: java is robust is robust java robust java is
/* * Enter a sentence and print it in rotation of words for all possible combinations. */ import java.io.*; class Rotation { public static void main (String args[])throws IOException { BufferedReader br=new BufferedReader(new InputStreamReader (System.in)); int i,j,p,x,l; String s; char ch; System.out.println("Enter a sentence terminated by a period(.) : "); s = br.readLine(); l = s.length(); String a[] = new String[l]; x=0; p=0; //Store the words in a string array for(i=0; i< l; i++){ ch = s.charAt(i); if(ch == ' ' || ch == '.'){ a[x++] = s.substring(p,i); p = i+1; } } System.out.println("\nOUTPUT:"); //Print the sentence in rotation of words for(i=0; i< x; i++){ for(j=i; j< x; j++){ System.out.print(a[j]+ " "); } for(j=0; j< i; j++){ System.out.print(a[j]+ " "); } System.out.println(); } }//end of main }//end of class
ISC Programs_1
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