poj2623

快排,注意保留一位小数。

View Code
#include <iostream>
#include
<cstdio>
#include
<cstdlib>
#include
<cstring>
#include
<algorithm>
#include
<cmath>
using namespace std;

#define maxn 250005

int n;
long long f[maxn], sum;

int main()
{
//freopen("t.txt", "r", stdin);
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf(
"%lld", &f[i]);
sort(f, f
+ n);
if (n & 1)
printf(
"%lld.0\n", f[n / 2]);
else
{
sum
= 0;
sum
+= f[n / 2];
sum
+= f[n / 2 - 1];
if (sum & 1)
printf(
"%lld.5\n", sum / 2);
else
printf(
"%lld.0\n", sum / 2);
}
return 0;
}

原文地址:https://www.cnblogs.com/rainydays/p/2047630.html