首页 > 代码库 > topcoder SRM 624 DIV2 BuildingHeightsEasy

topcoder SRM 624 DIV2 BuildingHeightsEasy

从大到小遍历一遍,每次取M个元素,然后求得最小的floor即可

    int minimum(int M, vector <int> heights)    {        sort(heights.begin(),heights.end());        int minFloor = 10000;        for(int i = heights.size()-1; i >=M-1; -- i){            int floor = 0;            for(int j = 0; j < M; ++j) floor +=heights[i]-heights[i-j];            minFloor =min(minFloor,floor);        }        return minFloor;    }