4.1.4--cup

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

1100 100 100 3141562

Sample Output

99.999024

解题思路：通过这题又了解了，精度太高也会超时，比如exp=1e-7如果换为1e-9就会超时

代码：

#include<iostream>#include<cmath>#include<cstdio>using namespace std;#define exp 1e-7const double pi=3.14159265;double r1,r2,h;double work(double &m){    double R=m*(r2-r1)/h+r1;    double v=(1.0/3.0)*pi*m*(R*R+r1*r1+R*r1);    return v;}int main(){    int t;    scanf("%d",&t);    double v,val;    while(t--)    {        scanf("%lf%lf%lf%lf",&r1,&r2,&h,&v);        double mid,minn=0,maxn=100;        while(minn<maxn)  //此处若为<=，也会超时，因为可能又一直循环的情况        {            mid=(minn+maxn)/2.0;            val=work(mid);            if(fabs(val-v)<exp) {break;}            else            {                if(val>v) maxn=mid;                else minn=mid;            }        }        printf("%.6lf\n",mid);    }    return 0;}