Day3-L-Cup HDU2289

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.

InputThe 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.

OutputFor 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

思路：可验证+单调性，二分答案，代码如下：

#define exp 1e-9

double r, R, V, H;

double calculateVolume(double height) {
    double nr = height / H * (R - r) + r;
    return acos(-1.0) / 3 * (r * r + r * nr + nr * nr) * height;
}




int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        scanf("%lf%lf%lf%lf", &r, &R, &H, &V);
        double left = 0.0, right = 100.0, mid;
        while(right - left > exp) {
            mid = (right + left) / 2;
            double tmp = calculateVolume(mid);
            if(fabs(tmp - V) <= exp)
                break;
            if(tmp > V)
                right = mid - exp;
            else
                left = mid + exp;
        }
        printf("%.6lf
", right);
    }
    return 0;
}

View Code

小结:带精度的问题都是上下限之差小于这个精度为while条件，(解方程和这一题