题目链接
https://www.patest.cn/contests/gplt/L2-015
思路 在求和的过程中 标记一下 最大值和最小值，在最后求平均的时候 用总和减去最大值和最小值 去除 （总数 - 2） 然后最后排序的时候 先按升序来排 然后 最后 POP 掉 多余的 再按升序来排
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <iostream> #include <algorithm> #include <cmath> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> using namespace std; typedef long long LL; const double PI = 3.14159265358979323846264338327; const double E = 2.718281828459; const double eps = 1e-6; const int MAXN = 0x3f3f3f3f; const int MINN = 0xc0c0c0c0; const int maxn = 1e6 + 5; const int MOD = 1e9 + 7; bool comp(double x, double y) { return x > y; } int main() { int n, k, m; scanf("%d%d%d", &n, &k, &m); vector <double> v; v.clear(); for (int i = 0; i < n; i++) { int Min = MAXN, Max = MINN; double sum = 0; int num; for (int j = 0; j < k; j++) { scanf("%d", &num); Min = min(Min, num); Max = max(Max, num); sum += num; } sum = (sum - Min - Max) * 1.0 / (k - 2); v.push_back(sum); } sort(v.begin(), v.end(), comp); while (v.size() > m) v.pop_back(); sort(v.begin(), v.end()); vector <double>::iterator it; int i; for (it = v.begin(), i = 0; it != v.end() && i < m; i++, it++) { if (i) printf(" "); cout << setprecision(3) << std::fixed << (*it); } }转载于:https://www.cnblogs.com/Dup4/p/9433265.html
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