zoukankan      html  css  js  c++  java
  • 1297

    1297 - Largest Box
    Time Limit: 2 second(s) Memory Limit: 32 MB

    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

    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

    题解:方程式列一下,因为是3次方,一看就是求极值的,但是注意范围是0,min(w,h)/2,三分写一下就过了。。。

    代码:

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<algorithm>
     6 #define mem(x,y) memset(x,y,sizeof(x))
     7 #define ZR (4*W+4*L)*(4*W+4*L)-48*W*L
     8 using namespace std;
     9 const int INF=0x3f3f3f3f;
    10 double L,W;
    11 double js(double x){
    12     return x*(L-2*x)*(W-2*x);
    13 }
    14 double sf(double l,double r){
    15     double mid,mm;
    16     while(r-l>1e-10){
    17         mid=(l+r)/2;
    18         mm=(mid+r)/2;
    19         if(js(mid)>=js(mm))r=mm;
    20         else l=mid;
    21     }
    22     return mid;
    23 }
    24 int main(){
    25     int T,flot=0;
    26     scanf("%d",&T);
    27     while(T--){
    28         scanf("%lf%lf",&L,&W);
    29         double x=sf(0,min(L,W)/2);
    30     //    printf("%f
    ",x);
    31         printf("Case %d: %lf
    ",++flot,js(x));
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    Cooperate with Myself
    A brief introduction of myself
    计算1+11+111+1111+........
    Jav实现F(n)=F(n-1)+F(n-2)+.....+F(1)+1
    查找二维数组中是否有符合的目标值
    排序算法
    时间复杂度
    Java代码实现单例模式
    查找一个字符串中重复出现字符的个数
    null,“”,empty的区别
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4947258.html
Copyright © 2011-2022 走看看