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Emmy von Rhoden was born on November 15, 1829, in Magdeburg, Germany.
1 answer
Emmy Rosenfeld has written:
'Neue Studien zur Lyrik von Friedrich von Spee'
1 answer
Max von der Gru n has written:
'Fahrtunterbrechung und andere Erza hlungen'
1 answer
/* write a program to print Days of Week using switch-case structure */
#include<stdio.h>
#include<conio.h>
void main()
{
int n;
clrscr();
printf("\n Enter Day of weak as Number 1 to 7 ");
scanf("%d",&n);
switch(n)
{
case 1:
printf("\n MONDAY ");
case 2:
printf("\n TUESDAY");
case 3:
printf("\n WEDNESDAY");
case 4:
printf("\n THURSDAY");
case 5:
printf("\n FRIDAY");
case 6:
printf("\n SATURDAY");
case 7:
printf("\n SUNDAY");
default :
printf("\n no operation is required");
}
getch();
}
1 answer
#include<iostream>
#include<conio.h> // for _getch()
void action1_1_1() { std::cout << "Performing action 1.1.1\n\n"; }
void action1_1_2() { std::cout << "Performing action 1.1.2\n\n"; }
void action1_1_3() { std::cout << "Performing action 1.1.3\n\n"; }
void action1_2_1() { std::cout << "Performing action 1.2.1\n\n"; }
void action1_2_2() { std::cout << "Performing action 1.2.2\n\n"; }
void action1_2_3() { std::cout << "Performing action 1.2.3\n\n"; }
void action1_3_1() { std::cout << "Performing action 1.3.1\n\n"; }
void action1_3_2() { std::cout << "Performing action 1.3.2\n\n"; }
void action1_3_3() { std::cout << "Performing action 1.3.3\n\n"; }
void action2_1_1() { std::cout << "Performing action 2.1.1\n\n"; }
void action2_1_2() { std::cout << "Performing action 2.1.2\n\n"; }
void action2_1_3() { std::cout << "Performing action 2.1.3\n\n"; }
void action2_2_1() { std::cout << "Performing action 2.2.1\n\n"; }
void action2_2_2() { std::cout << "Performing action 2.2.2\n\n"; }
void action2_2_3() { std::cout << "Performing action 2.2.3\n\n"; }
void action2_3_1() { std::cout << "Performing action 2.3.1\n\n"; }
void action2_3_2() { std::cout << "Performing action 2.3.2\n\n"; }
void action2_3_3() { std::cout << "Performing action 2.3.3\n\n"; }
void action3_1_1() { std::cout << "Performing action 3.1.1\n\n"; }
void action3_1_2() { std::cout << "Performing action 3.1.2\n\n"; }
void action3_1_3() { std::cout << "Performing action 3.1.3\n\n"; }
void action3_2_1() { std::cout << "Performing action 3.2.1\n\n"; }
void action3_2_2() { std::cout << "Performing action 3.2.2\n\n"; }
void action3_2_3() { std::cout << "Performing action 3.2.3\n\n"; }
void action3_3_1() { std::cout << "Performing action 3.3.1\n\n"; }
void action3_3_2() { std::cout << "Performing action 3.3.2\n\n"; }
void action3_3_3() { std::cout << "Performing action 3.3.3\n\n"; }
void submenu1_1()
{
while (true)
{
std::cout
<< "SUBMENU 1.1\n===========\n\n"
<< "1 - action 1.1.1\n"
<< "2 - action 1.1.2\n"
<< "3 - action 1.1.3\n"
<< "X - exit to submenu 1\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action1_1_1(); break;
case '2': action1_1_2(); break;
case '3': action1_1_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu1_2()
{
while (true)
{
std::cout
<< "SUBMENU 1.2\n===========\n\n"
<< "1 - action 1.2.1\n"
<< "2 - action 1.2.2\n"
<< "3 - action 1.2.3\n"
<< "X - exit to submenu 1\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action1_2_1(); break;
case '2': action1_2_2(); break;
case '3': action1_2_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu1_3()
{
while (true)
{
std::cout
<< "SUBMENU 1.3\n===========\n\n"
<< "1 - action 1.3.1\n"
<< "2 - action 1.3.2\n"
<< "3 - action 1.3.3\n"
<< "X - exit to submenu 1\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action1_3_1(); break;
case '2': action1_3_2(); break;
case '3': action1_3_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu2_1()
{
while (true)
{
std::cout
<< "SUBMENU 2.1\n===========\n\n"
<< "1 - action 2.1.1\n"
<< "2 - action 2.1.2\n"
<< "3 - action 2.1.3\n"
<< "X - exit to submenu 2\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action2_1_1(); break;
case '2': action2_1_2(); break;
case '3': action2_1_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu2_2()
{
while (true)
{
std::cout
<< "SUBMENU 2.2\n===========\n\n"
<< "1 - action 2.2.1\n"
<< "2 - action 2.2.2\n"
<< "3 - action 2.2.3\n"
<< "X - exit to submenu 2\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action2_2_1(); break;
case '2': action2_2_2(); break;
case '3': action2_2_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu2_3()
{
while (true)
{
std::cout
<< "SUBMENU 2.3\n===========\n\n"
<< "1 - action 2.3.1\n"
<< "2 - action 2.3.2\n"
<< "3 - action 2.3.3\n"
<< "X - exit to submenu 2\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action2_3_1(); break;
case '2': action2_3_2(); break;
case '3': action2_3_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu3_1()
{
while (true)
{
std::cout
<< "SUBMENU 3.1\n===========\n\n"
<< "1 - action 3.1.1\n"
<< "2 - action 3.1.2\n"
<< "3 - action 3.1.3\n"
<< "X - exit to submenu 3\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action3_1_1(); break;
case '2': action3_1_2(); break;
case '3': action3_1_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu3_2()
{
while (true)
{
std::cout
<< "SUBMENU 3.2\n===========\n\n"
<< "1 - action 3.2.1\n"
<< "2 - action 3.2.2\n"
<< "3 - action 3.2.3\n"
<< "X - exit to submenu 3\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action3_2_1(); break;
case '2': action3_2_2(); break;
case '3': action3_2_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu3_3()
{
while (true)
{
std::cout
<< "SUBMENU 3.3\n===========\n\n"
<< "1 - action 3.3.1\n"
<< "2 - action 3.3.2\n"
<< "3 - action 3.3.3\n"
<< "X - exit to submenu 3\n\n";
int i = _getch();
switch ((char) i)
{
case '1': action3_3_1(); break;
case '2': action3_3_2(); break;
case '3': action3_3_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu1()
{
while (true)
{
std::cout
<< "SUBMENU 1\n=========\n\n"
<< "1 - sub-menu 1.1\n"
<< "2 - sub-menu 1.2\n"
<< "3 - sub-menu 1.3\n"
<< "X - exit to main menu\n\n";
int i = _getch();
switch ((char) i)
{
case '1': submenu1_1(); break;
case '2': submenu1_2(); break;
case '3': submenu1_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu2()
{
while (true)
{
std::cout
<< "SUBMENU 2\n=========\n\n"
<< "1 - sub-menu 2.1\n"
<< "2 - sub-menu 2.2\n"
<< "3 - sub-menu 2.3\n"
<< "X - exit to main menu\n\n";
int i = _getch();
switch ((char) i)
{
case '1': submenu2_1(); break;
case '2': submenu2_2(); break;
case '3': submenu2_3(); break;
case 'X':
case 'x': return;
}
}
}
void submenu3()
{
while (true)
{
std::cout
<< "SUBMENU 3\n=========\n\n"
<< "1 - sub-menu 3.1\n"
<< "2 - sub-menu 3.2\n"
<< "3 - sub-menu 3.3\n"
<< "X - exit to main menu\n\n";
int i = _getch();
switch ((char) i)
{
case '1': submenu3_1(); break;
case '2': submenu3_2(); break;
case '3': submenu3_3(); break;
case 'X':
case 'x': return;
}
}
}
int main()
{
while (true)
{
std::cout
<< "MAIN MENU\n=========\n\n"
<< "1 - sub-menu 1\n"
<< "2 - sub-menu 2\n"
<< "3 - sub-menu 3\n"
<< "X - exit program\n\n";
int i = _getch();
switch ((char) i)
{
case '1': submenu1(); break;
case '2': submenu2(); break;
case '3': submenu3(); break;
case 'X':
case 'x': return 0;
}
}
}
1 answer
The plural of the proper noun Emmy is usually Emmys, and very rarely Emmies. This is a trademark name, and therefore distinct from abbreviations that use upper and lower case. It is also seen using an apostrophe, but this is technically erroneous.
To avoid having to decide, you can always use "Emmy awards."
1 answer
The sum from 1 to n is n*(n+1)/2
In this case that mean 20*21/2 = 210
The sum from 1 to n is n*(n+1)/2
In this case that mean 20*21/2 = 210
The sum from 1 to n is n*(n+1)/2
In this case that mean 20*21/2 = 210
The sum from 1 to n is n*(n+1)/2
In this case that mean 20*21/2 = 210
2 answers
The cast of Im Busch von Mexiko - 1967 includes: Anita Bucher as Stimme Thea Grodtzinsky Paul Henckels Emmy Julich as Hedwig Maier Gerhart Lippert as Rundfunksprecher
1 answer
- #include
- main()
- {
- char ch;
- clrscr();
- pritnf("\nEnter any character:");
- fflush(stdin);
- scanf("%c",&ch);
- if (ch 'U' ch = 'u')
- {
- printf("\n The entered character %c is Vowel.\n", ch);
- }
- else
- {
- printf("\nThe entered character %c is not Vowel.\n",ch);
- }
- }
5 answers
Best case for insertion sort is O(n), where the array is already sorted. The worst case, where the array is completely reversed, is O(n*n).
3 answers
The best and worst case time complexity for heapsort is O(n log n).
1 answer
There is no worst case for merge sort. Each sort takes the same amount of steps, so the worst case is equal to the average case and best case. In each case it has a complexity of O( N * log(N) ).
5 answers
nX in this case this says that n multiplys X by a factor of n
Xn in this case this says that X is to the power of factor n
not really another other case I can think of. Not sure what you are asking
1 answer
The cast of Die verwunschene Prinzessin - 1919 includes: Ernst Behmer Olga Dalzell Olga Engl Eva May Johannes Riemann Leopold von Ledebur Anna von Palen Emmy Wyda
1 answer
The cast of Krawattenmacher - Der Wucherer von Berlin - 1922 includes: Ida Andorffy Emmy Denner Richard Franz Stavy Greder Margarete Kupfer Rolf Loer Leo Sloma
1 answer
Merge sort is O(n log n) for both best case and average case scenarios.
2 answers
The upper case N has no line of symmetry because if you cut the N in half in any ways it will still not be symmetrical.
1 answer
The best case time complexity of heapsort is O(n log n), where n is the number of elements in the input array.
1 answer
The worst case time complexity of heapsort is O(n log n), where n is the number of elements in the input array.
1 answer
Emmy is a nickname for the image orthicon tube used in television cameras.
1 answer
#include <stdio.h>
int main()
{
int month,date;
printf("Enter the month and date of your birth in the format mm-dd:\n");
scanf("%d-%d",&month,&date);
if(month<1month>12date<1date>31) {
printf("That is not a valid date.\n");
return 1;
}
printf("Your zodiac sign is ");
switch(month) {
case 1:
if (date<20) printf("Capricorn\n");
else printf("Aquarius\n");
break;
case 2:
if (date<19) printf("Aquarius\n");
else printf("Pisces\n");
break;
case 3:
if (date<21) printf("Pisces\n");
else printf("Aries\n");
break;
case 4:
if (date<20) printf("Aries\n");
else printf("Taurus\n");
break;
case 5:
if (date<21) printf("Taurus\n");
else printf("Gemini\n");
break;
case 6:
if (date<21) printf("Gemini\n");
else printf("Cancer\n");
break;
case 7:
if (date<23) printf("Cancer\n");
else printf("Leo\n");
break;
case 8:
if (date<23) printf("Leo\n");
else printf("Virgo\n");
break;
case 9:
if (date<23) printf("Virgo\n");
else printf("Libra\n");
break;
case 10:
if (date<23) printf("Libra\n");
else printf("Scorpio\n");
break;
case 11:
if (date<22) printf("Scorpio\n");
else printf("Sagittarius\n");
break;
case 12:
if (date<22) printf("Sagittarius\n");
else printf("Capricorn\n");
break;
}
return 0;
}
1 answer
If it is an unbalanced binary tree, O( ln( n ) / ln( 2 ) ) is best-case. Worst case is O( n ).
If it is balanced, worst case is O( ln( n ) / ln( 2 ) ).
1 answer
The worst case occurs when data is already sorted where the complexity is O(n^2) instead of the well known O(n log n)
1 answer
The worst case time complexity of heap sort is O(n log n), where n is the number of elements in the input array.
1 answer
here's my code to count the letters in a phrase...can anyone tell me what modification i should make in this program ...to make it count the word size and diaplaying according to the word size...
#include
o o 1 0 2 5 3 2 4 1 and so on... it should display according to the word size...
1 answer
