HDU 2289 Cup

C - Cup

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeHDU 2289

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 

 
二分猜数字类型的题
这类题到现在做的也不少了
基本方法就是，给定区间最大最小值，在这个区间中二分出来一个mid值，然后判断这个mid值与最优解的大小关系，从而调整区间范围。直至找到满足要求的最终解。
刘汝佳在小白书中把判断mid的过程称为构造谓词P(x)函数。做的越多也就理解的越深刻。

#include <stdio.h>#define EPS 1e-8#define PI 3.141592653double r, R, H, V;double cal(double r,double R,double h,double H){    double u = h/H*(R-r) + r;    return PI/3*(r*r+r*u+u*u)*h;}int main(){        int t;        scanf("%d",&t);        while(t--)        {                scanf("%lf%lf%lf%lf",&r,&R,&H,&V);                double minn=0,maxn=100,mid;                while(maxn-minn>EPS)                {                        mid=(maxn+minn)/2;                        double cmid=cal(r,R,mid,H);                        if(cmid<V)                                minn=mid;                        else                                maxn=mid;                }                printf("%.6lf\n",mid);        }        return 0;}

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