Problem G
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 98 Accepted Submission(s) : 27
Problem Description
The WHU ACM Team has a big cup, with which every member drinks water. Now, we know the volume of the water in the cup, can you tell us it height? The radius of the cup's top and bottom circle is known, the cup's height is also known.
Input
The input consists of several test cases. The first line of input contains an integer T, indicating the num of test cases. Each test case is on a single line, and it consists of four floating point numbers: r, R, H, V, representing the bottom radius, the top radius, the height and the volume of the hot water. Technical Specification 1. T ≤ 20. 2. 1 ≤ r, R, H ≤ 100; 0 ≤ V ≤ 1000,000,000. 3. r ≤ R. 4. r, R, H, V are separated by ONE whitespace. 5. There is NO empty line between two neighboring cases.
Output
For each test case, output the height of hot water on a single line. Please round it to six fractional digits.
Sample Input
1 100 100 100 3141562
Sample Output
99.999024
总结：这道题做了很多次，反正每次都是因为一些小问题出现错误
1）输入数据时候，不需要用while（~scanf("%d",&t)）语句，原因是题目已知测试组数
2）题目中水的体积计算公式：
圆柱体的体积：V=pi*(r*r+R*R+r*R)/3.0(其中：pi #define pi 4.0*atan(1.0);
r,R:两个底半径）
3）此题采用二分法：以水的高度二分
设水的高度已知，为hx，那么水高的取值为【0,100】
4）在GCC的编译环境中，要将printf("%.4lf\n",mid);
变为printf("%.4f\n",mid);
代码如下：
#include<stdio.h> #include<math.h> #define pi 4.0*atan(1.0) #define eps 1e-9 double r,R,H,V; double f(double h) { double k=h/H*(R-r)+r; return pi*h*(k*k+r*r+r*k)/3.0; } int main() { int t; double left,right,mid; scanf("%d",&t); while(t--) { scanf("%lf%lf%lf%lf",&r,&R,&H,&V); left=0.0; right=100.0; while(right-left>eps) { mid=(right+left)/2.0; if(f(mid)>V) right=mid; else left=mid; } printf("%.6f\n",mid); } return 0; }