Pie Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7639 Accepted: 2817 Special Judge 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. 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. 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. Sample Input 3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2 Sample Output 25.1327 3.1416 50.2655 Source Northwestern Europe 2006 愁啊，這精度問題真是讓人不知所措啊，思路正確因為精度問題卡在那了，最初做題的時候double數據強轉成int， 直接轉，後來隨著做題，因為在強轉的時候沒有加0.01錯了很多次，所以就養成了，強轉是+0.01的習慣，而今天做的這個題目是必須不能加，加了就錯，直接強轉就可以了。 [cpp] #include <stdio.h> #include <string.h> #include <math.h> #define PI 3.1415926535898 #define EQS 1e-7 double a[11000]; int n,m; int main() { double binary_search(double l,double r); int t,i; double res,max; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&m); m++; for(i=0,max=0;i<=n-1;i++) { scanf("%lf",&a[i]); if(a[i]>max) { max=a[i]; } } res=binary_search(0,max*max*PI); printf("%.4lf\n",res); } return 0; } int check(double mid) { int s=0,i; for(i=0;i<=n-1;i++) { s+=(int)(a[i]*a[i]*PI/mid); } if(s>=m) { return 1; }else { return 0; } } double binary_search(double l,double r) { int t; double mid; while(!(fabs(r-l)<=EQS)) { mid=(r+l)/2.0; if(!check(mid)) { r=mid; }else { l=mid; } } return l; }
