LightOJ 1297

 LightOJ 1297 

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice

Description

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.

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 than 10^-6 will be ignored.

Sample Input

3

2 10

3.590 2.719

8.1991 7.189

Sample Output

Case 1: 4.513804324

Case 2: 2.2268848896

Case 3: 33.412886

题意：在矩形的四个角割掉四个边长为X的小正方形，求出剩下的部分折成的无顶长方体or（正方体）的体积最大为？？？

    思路：V=x*（w-2*x）*（l-2x） ，然后对其进行求导得到变化趋势，再用二分求得最大值!

 代码：

      

#include<stdio.h>#include<string.h>double w,l;double min(double x,double y){if(x>y)return y;return  x;}double  js(double  x){return  x*(w-2*x)*(l-2*x);//体积的计算！ }double  jsd(double  x){return (w-2*x)*(l-2*x)-2*x*(l-2*x)-2*x*(w-2*x);//倒数！ }int main(){int t,mark=1;scanf("%d",&t);while(t--){scanf("%lf%lf",&w,&l);double  left=0.0,right=min(w,l),mid;while(right-left>=0.00000001)//二分 { mid=(left+right)/2;double  count=jsd(mid);if(count>0)//在递增的区间倒数大于零！ left=mid;elseright=mid;}printf("Case %d: %.6lf\n",mark++,js(mid));}return  0;}