LightOJ 1297 - Largest Box (二分查找)



1297 - Largest Box

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Time Limit: 2 second(s)

Memory Limit: 32 MB

In the following figure you can see a rectangular card. The width of the card isW 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 ofx.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing two real numbers L and W (0 < L, W < 100).

Output

For each case, print the case number and the maximum volume of the box that can be made. Errors less than10^-6 will be ignored.

Sample Input

Output for Sample Input

3

2 10

3.590 2.719

8.1991 7.189

Case 1: 4.513804324

Case 2: 2.2268848896

Case 3: 33.412886

 

题意：给出一纸片，长为L，宽为W，如图将其减去4个边长为x的正方形，在按虚线上折，成为一个没有盖子的长方体容器。x的大小是不确定的，问容器最大容积是多少

先得到长方体的体积式，再二分查找最大体积，代码如下：

#include<cstdio>#include<cmath>double l,w;double g(double x){return 12*x*x-4*(w+l)*x+w*l;}double f(double x){return 4*x*x*x-2*(w+l)*x*x+w*l*x;}int main(){int t,k=1;double z,left,right,mid;scanf("%d",&t);while(t--){scanf("%lf%lf",&l,&w);if(l>w)z=w;else z=l;left=0;right=z/2;while(right-left>1e-8){mid=(left+right)/2;if(g(mid)<0)right=mid;elseleft=mid;}printf("Case %d: %.6lf\n",k++,f(mid));}return 0;}