Educational Codeforces Round 49 (Rated for Div. 2) ---- C. Minimum Value Rectangle （贪心）

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题目链接

做法：数组不要开超过1e4否则会超时……

自己真的要好好改改硬解题这个坏毛病……

我们设长和宽分别为a,b

P^2/S = (2*(a+b))^2/a*b = 4(a/b+b/a+2)

也就是说当a/b+b/a 最小的时候，P^2/S 最小，然后简单贪心处理一下就好了。

坑点就在数组开太大，就会超时 ORZ

AC代码：

#include<bits/stdc++.h>
#define rep(i,s,t) for(int i = (int)(s); i <= (int)(t); i++)
#define rev(i,t,s) for(int i = (int)(t); i >= (int)(s); i--)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define sz(x) (int)(x).size()
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxn = 1e4+5;
const int INF = 0x3f3f3f3f;
int e[maxn];
int vis[maxn];
inline int read()
{
    char x;
    while((x = getchar())<'0' || x>'9');
    int u = x-'0';
    while((x = getchar())>='0' && x<='9') u = (u<<3)+(u<<1)+x-'0';
    return u;
}
int main()
{
    #ifdef LOCAL_FILE
    freopen("in.txt","r",stdin);
    #endif // LOCAL_FILE
    int t;
    t = read();
    while(t--)
    {
        memset(e,0,sizeof(e));
        memset(vis,0,sizeof(vis));
        int n,d;
        int flag = 1;
        n = read();
        int k=0;
        for(int i=0;i<n;i++)
        {
            d = read();
            vis[d]++;
            if(vis[d] == 4 && flag)
            {
                printf("%d %d %d %d\n",d,d,d,d);
                flag = 0;
            }
            if(vis[d] == 2 && flag)
                e[k++] = d;
        }
        if(flag)
        {
            sort(e,e+k);
            double mmin = INF;
            int ans1,ans2;
            for(int i=0;i<k-1;i++)
            {
                double tmp = (double)e[i]/e[i+1]+(double)e[i+1]/e[i];
                if(tmp<mmin)
                {
                    mmin = tmp;
                    ans1 = e[i];
                    ans2 = e[i+1];
                }
            }
            printf("%d %d %d %d\n",ans1,ans1,ans2,ans2);
        }
    }
    return 0;
}