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);
}
}