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Through the use of computational thinking techniques, models and algorithms can be created. A(n) can be created that represents the generalized problem. A(n) , or set of step-by-step instructions, can be developed to solve the problem words can be used is
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Computers and Technology, 21.06.2019 13:30
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Computers and Technology, 21.06.2019 18:30
Ayear in the modern gregorian calendar consists of 365 days. in reality, the earth takes longer to rotate around the sun. to account for the difference in time, every 4 years, a leap year takes place. a leap year is when a year has 366 days: an extra day, february 29th. the requirements for a given year to be a leap year are: the year must be divisible by 4if the year is a century year (1700, 1800, the year must be evenly divisible by 400some example leap years are 1600, 1712, and 2016.write a program that takes in a year and determines whether that year is a leap year.ex: if the input is 1712, the output is: 1712 is a leap year.ex: if the input is 1913, the output is: 1913 is not a leap year.
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Computers and Technology, 22.06.2019 13:00
Write a program which asks you to enter a name in the form of first middle initial last. so you might enter for example samuel p. clemens. use getline to read in the string because it contains spaces. also, apparently the shift key on your keyboard doesn’t work, because you enter it all lower case. pass the string to a function which uses .find to locate the letters which need to be upper case and use toupper to convert those characters to uppercase. the revised string should then be returned to main in the form last, first mi where it will be displayed.
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Computers and Technology, 22.06.2019 20:00
What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }
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