Topcoder SRM 633 div.2 500 jumping

Source: Internet
Author: User

Question: Give a vertex (x, y), give some step delta1, delta2... deltan: Can I reach the (x, y) point after completing n steps in strict accordance with the step (0, 0.

Solution: in fact, it is to determine whether these line segments and (0, 0)-(x, y) can constitute a multilateral (angle ?) You only need to judge whether the longest side is less than half of the length of all sides.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;class Jumping{public:    string ableToGet(int x, int y, vector <int> jumpLengths)    {        double dis = sqrt(x*x+y*y);        double sum = dis,maxi = dis;        for(int i=0;i<jumpLengths.size();i++)        {            sum += (double)jumpLengths[i];            maxi = max(maxi,(double)jumpLengths[i]);        }        if(sum-maxi >= maxi)            return "Able";        return "Not able";    }};
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Topcoder SRM 633 div.2 500 jumping

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