One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment. Vitaly sees a building of n floors and 2· m windows on each floor. On each floor there are m flats numbered from 1 to m , and two consecutive windows correspond to each flat. If we number the windows from 1 to 2· m from left to right, then the j -th flat of the i -th floor has windows 2· j - 1 and 2· j in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on. Given the information about the windows of the given house, your task is to calculate the number of flats where, ac...
A. Turtle Puzzle: Rearrange and Negate time limit per test 2 seconds memory limit per test 256 megabytes You are given an array a � of n � integers. You must perform the following two operations on the array (the first, then the second): Arbitrarily rearrange the elements of the array or leave the order of its elements unchanged. Choose at most one contiguous segment of elements and replace the signs of all elements in this segment with their opposites. Formally, you can choose a pair of indices l , r � , � such that 1 ≤ l ≤ r ≤ n 1 ≤ � ≤ � ≤ � and assign a i = − a i � � = − � � for all l ≤ i ≤ r � ≤ � ≤ � (negate elements). Note that you may choose not to select a pair of indices and leave all the signs of the elements unchanged. What is the maximum sum of the array elements after performing these two operations (the first, then the second)? Input The first line of the input contains a single int...
Problem link Solution O(1) #include <stdio.h> #include <iostream> using namespace std ; int main () { long int l , r ; int t ; cin >> t ; long int x = l , y = r ; while ( t --) { scanf ( "%ld" ,& l ); scanf ( "%ld" ,& r ); y = r ; int mod =( r % l ); cout << l << " " << r - mod << endl ; mod = 0 ; } return 0 ;}
Comments
Post a Comment