[topcoder]SRM 633 DIV 2

第一题，http://community.topcoder.com/stat?c=problem_statement&pm=13462&rd=16076

模拟就可以了。

#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;
  }
};

第二题，想了很久。最后发现用三角形的A+B>=C，一个一个推，可以推出N条边所组成的多边形（开口）的距离范围。http://apps.topcoder.com/wiki/display/tc/SRM+633#Jumping

有详细的图示。

#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";
    }

  }

};

第三题，没做。后来看题解，就是用LCD和GCD的限制，得到x*y，然后穷举搜索。用DFS。