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Cup
Cup |
| Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 932 Accepted Submission(s): 319 |
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 |
Source The 4th Baidu Cup final |
Recommend lcy |
/*一晚上把底半径和二分的右区间变量名用重了，还看出来，真是.....*/#include<bits/stdc++.h>#define op 1e-9#define pi acos(-1.0)using namespace std;/*V=π×h×(R2﹢R×r﹢r2)/3 */double S(double r,double R,double h,double H)//圆台面积{ double r1=r+h*(R-r)/H; return pi*h*(r1*r1+r1*r+r*r)/3; }int main(){ //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin); int t; double left,right,h,v; double r,R,H,V; scanf("%d",&t); while(t--) { scanf("%lf%lf%lf%lf",&r,&R,&H,&V); //cout<<r<<" "<<R<<" "<<H<<" "<<(int )V<<endl; left=0,right=100; while(right-left>op) { //cout<<h<<endl; h=(right+left)/2; v=S(r,R,h,H); //cout<<v<<endl; if(fabs(v-V)<=op) break; else if(v>V) right=h-op; else left=h+op; } printf("%.6lf\n",h); } return 0;}
Cup
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