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Problem Description
My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.
My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size. What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different. |
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Input
One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends. ---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies. |
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Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).
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Sample Input
3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2 |
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Sample Output
25.1327 3.1416 50.2655 |
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Codes
1 #include <stdio.h> 2 #include <math.h> 3 #define N 10000 4 #define PI acos(-1.0) 5 int n, f; 6 int ri[N + 5]; 7 double v[N + 5]; 8 //参数是体积，判断该体积下，可以分到蛋糕的人数是否大于等于指定人数 9 //parameter: volume 10 //function: judge if the all the people get the cake in the case of the volume 11 int isSuitable(double vol) 12 { 13 int num, i; 14 num = 0; 15 for (i = 1; i <= n; i++) 16 { 17 num += (int)(v[i] / vol); 18 } 19 if (num < f) 20 return 0; 21 else 22 return 1; 23 } 24 void main() 25 { 26 //declare 27 int i, j, count; 28 double min, max, mid, sum, res; 29 sum = 0; 30 scanf("%d", &count); 31 for (j = 1; j <= count; j++) 32 { 33 //input 输入 34 scanf("%d %d", &n, &f); 35 for (i = 1; i <= n; i++) 36 { 37 scanf("%d", &ri[i]); 38 v[i] = ri[i] * ri[i]; 39 sum += v[i]; 40 } 41 //dichotomy 二分法-------------------- 42 f += 1; 43 min = 0; 44 max = sum / f; 45 //如果最大值都满足条件，那么就答案就是这个最大值 46 //firstly, judge if the max is suitable 47 if (isSuitable(max)) 48 res = max; 49 else 50 { 51 //find the the result by using dictomoy 52 while (max - min > 1e-6) 53 { 54 mid = (max + min) / 2; 55 if (isSuitable(mid)) 56 min = mid; 57 else 58 max = mid; 59 } 60 } 61 //output 62 printf("%.4f\n", min * PI); 63 } 64 } |
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Key Points
Firstly, in order to gain high precision, we must use "acos(-1.0)", which need add "#include <math.h>" instand of "3.1415926", and write "1e-6" instand of "1e-5".
Secondly, there is no "bool" in C, but there is in C++.
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