In tim O'Brien's "Ambush," ehich element directly enhances the theme of the pasts imprint on the present

When keyboarding, if you want to be fast and accurate you should return your hands on home row with your fingers relaxed and your eyes on the text use only your index fingers to hunt and peck keep your fingers stiff and your eyes on the keys slouch in your chair and keep your fingers hovering over all of the keys

You will be given two character arrays of the same size, one will contain a number of ships. ships will move around the character array based on which way they are facing and the route they are on. routes are given in the other array. the route consists of '-' and '|' for straight paths, '\' and '/' for curves, and '+' for intersections. there are ships on these routes. ships always face a direction, '^' for up, '> ' for right, 'v' for down, and '< ' for left. any time the ships hit a '\' or a '/' it will turn as you would expect a ship to turn (e.g. a '^' that moves into a '/' will turn right). at an intersection, ships will always continue straight through. all ships move at the same speed, ships take turns moving and all ships move during one 'tick'. the one in the most top left goes first, followed by those to its right, then the ones in the next row. it iterates along the rows and then down the columns. each ship moves one space on its turn moving along the route. your function needs to return the position of the first collision between two ships and the number of ticks before the crash occurred.

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; }