hdu 2289 Cup (二分法)
Cup
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3533 Accepted Submission(s): 1136
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.
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.
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
Source
Recommend
错了好多次.主要是求体积的公式没找对
1 //0MS 228K 573 B G++ 2 #include<stdio.h> 3 #include<math.h> 4 #define pi acos(-1.0) 5 #define exp 1e-10 6 double V(double r1,double r2,double h) 7 { 8 9 return h*pi*(r1*r1+r2*r2+r1*r2)/3; 10 } 11 int main(void) 12 { 13 int t; 14 double r1,r2,h,v; 15 scanf("%d",&t); 16 while(t--) 17 { 18 scanf("%lf%lf%lf%lf",&r1,&r2,&h,&v); 19 double l=0,r=h; 20 while(r-l>exp){ 21 double mid=(r+l)/2; 22 double tr2=(r2-r1)*mid/h+r1; 23 if(V(r1,tr2,mid)>v) r=mid; 24 else l=mid; 25 } 26 printf("%.6lf\n",l); 27 } 28 return 0; 29 }
