LightOJ

链接：

题意：

In the following figure you can see a rectangular card. The width of the card is W and length of the card is L and thickness is zero. Four (x*x) squares are cut from the four corners of the card shown by the black dotted lines. Then the card is folded along the magenta lines to make a box without a cover.

Given the width and height of the box, you will have to find the maximum volume of the box you can make for any value of x.

思路：

三次函数。
求导，根据图像求二次函数的较小x解即可。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>

using namespace std;
typedef long long LL;
const int INF = 1e9;

const int MAXN = 1e6+10;
const int MOD = 1e9+7;

int main()
{
    int t, cnt = 0;
    double l, w;
    scanf("%d", &t);
    while(t--)
    {
        printf("Case %d: ", ++cnt);
        scanf("%lf%lf", &l, &w);
        double res = (4*(l+w)-sqrt(16*(l+w)*(l+w)-4*12*(l*w)))/24.0;
        printf("%lf
", res*(l-2*res)*(w-2*res));
    }

    return 0;
}