| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10309 | Accepted: 3651 | 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. 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
Output
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
AC代码:
#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
double pi=3.14159265359; //精度要足够大啊,要不wa死了
double v[10010];
int n,m;
double s,vm;
int judge(double mid){
int sum=0;
for(int i=1;i<=n;i++)
sum+=(int)(v[i]/mid);
if(sum>=m)
return 1;
return 0;
}
int main(){
int T; scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
s=0.0; m++;
for(int i=1;i<=n;i++){
double r; scanf("%lf",&r);
v[i]=pi*r*r;
s+=v[i];
}
vm=s/(m+0.0);
double left,right,mid;
left=0.0; right=vm;
while((right-left)>0.000001){
mid=(right+left)/2;
if(judge(mid))
left=mid;
else
right=mid;
}
printf("%.4lf\n",left);
}
return 0;
}
原文:http://www.cnblogs.com/gcczhongduan/p/5132575.html