Question

In class we saw that the fast divide and conquer integer multiplication method takesO(nlog3) time. Thisis because the the shifts and additions can be done in time O(n). Suppose we use a very inefficient method for doingthe shifts and additions, so these take timeO(n2). With this slower method for doing the shifts and additions, figureout how much time will the divide and conquer integer multiplication method take . You can assume that nothing elsechanges in the divide and conquer integer multiplication method.(a) What will be the new recurrence relation for divide and conquer integer multiplication, if we are using the inefficientmethod for doing the shifts and additions(b) What will be the solution to the new recurrence relation.

Answer 1

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In general or more specifically in competitive programming there are many instances where we need to convert a number to a string or string to a number. But lack of knowledge of certain essential tools bind us to do so. Some methods to achieve this task are mentioned in this article.

Converting string to number

Method 1 : Using stringstream class or sscanf()
Method 2 : String conversion using stoi() or atoi()
Both these methods have been discussed in detail in the this article.

Method 3 : Using boost lexical cast
Boost library offers an inbuild function “lexical_cast(“string”)”, which directly converts a string to number. It returns an exception “bad_lexical_cast” in case of invalid input.

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//C++ code to demonstrate working of lexical_cast() #include<iostream> #include <boost/lexical_cast.hpp>// for lexical_cast() #include <string> // for string using namespace std; int main() {    string str = "5";    string str1 = "6.5";    // Initializing f_value with casted float    // f_value is 6.5    float f_value = boost::lexical_cast<float>(str1);    // Initializing i_value with casted int    // i_value is 5    int i_value = boost::lexical_cast<int>(str);

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