Math With Styeps
vector numOfBurgers(int ts, int cs) {\n        vector ans;\n        /\n        x4 + y2 = ts    (1)\n        x1 + y1 = cs    (2)\n            \n        2(x2 + y)
vector<int> numOfBurgers(int ts, int cs) {\n        vector<int> ans;\n        /*\n        x*4 + y*2 = ts    (1)\n        x*1 + y*1 = cs    (2)\n            \n        2(x*2 + y) = ts\n        x*1 + y*1 = cs\n        \n        divide (1) by 2  \n        x*2 +y = ts/2   (3)\n        x + y = cs\n            \n\t\t(3)-(2) yields\n        x = ts/2 -cs\n\t\t\n        subsitute into (2) and derive for y\n        y =cs -x\n        */\n        \n\t\t//if ts is odd then there is not solution, from (1)\n        if(ts%2)return ans;\n        \n        int x= ts/2-cs;\n        int y = cs-x;\n        \n\t\t//x and y can not be zero in this context\n        if(x<0 || y<0)return ans;\n        \n        ans.push_back(x);\n        ans.push_back(y);\n        \n        return ans;\n    }
['Math', 'C']