[TopCoder] SRM 633 DIV 2

First question, http://community.topcoder.com/stat?c=problem_statement&pm=13462&rd=16076.

The simulation is ready.

#include <vector> #include <algorithm>using namespace Std;class Target {public:  vector <string> Draw (int n) {    vector<string> result (n, String (n, '));    int x = 0;    int y = 0;    while (n >= 1) {for      (int i = 0; i < n; i++) {        result[x + i][y] = ' # ';        Result[x][y + i] = ' # ';        Result[x + n-1][y + i] = ' # ';        Result[x + i][y + n-1] = ' # ';      }      x + = 2;      Y + = 2;      N-= 4;    }    return result;}  ;

The second question, thought for a long time. Finally, we find that the a+b>=c of the triangle, a push, can introduce the distance range of the polygon (opening) composed of n edges. Http://apps.topcoder.com/wiki/display/tc/SRM+633#Jumping

There are detailed illustrations.

#include <vector> #include <algorithm> #include <string> #include <algorithm>using namespace Std;class Jumping {public:  string abletoget (int x, int y, vector <int> jumplengths) {    double d = sqrt (1.0 * X * x + 1.0 * y * y);    Sort (Jumplengths.begin (), Jumplengths.end ());    Double low = jumplengths[0];    Double high = jumplengths[0];    for (int i = 1; i < jumplengths.size (); i++) {Low      = max (0.0, Jumplengths[i] – high);      High = high + jumplengths[i];    }    if (d >= low && D <= High) {      return ' Able ';    } else {      return ' not Able ';    }}  ;

The third question, did not do. Later see the puzzle, is to use the LCD and gcd restrictions, get x*y, and then exhaustive search. With DFS.

[TopCoder] SRM 633 DIV 2