poj3122 二分答案+贪心
2017-06-02 19:25
477 查看
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.
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.
#include<stdio.h> const double pi=3.1415926535897932; double sum; int cnt,n,f,T; double p[10005]; inline char nc(){ static char buf[100000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; } inline int read(){ register int x=0,f=1; register char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f*=-1;ch=getchar();} while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar(); return f*x; } int main(){ T=read(); while(T--){ sum=0,n=read(),f=read(); f++; for(register int i=1;i<=n;i++){ p[i]=read(); p[i]=p[i]*p[i]; sum+=p[i]; } double lf=0.0,rg=sum/f,mid; while(rg-lf>0.00001){ mid=(lf+rg)/2; cnt=0; for(register int i=1;i<=n;i++) cnt+=(int)(p[i] / mid); if(cnt>=f) lf=mid; else rg=mid; } printf("%0.4lf\n",mid*pi); } }
相关文章推荐
- [POJ 3122] Pie 二分答案+贪心
- POJ3122-Pie-二分答案
- POJ 3122(二分答案)
- Poj 3122 Pie 二分+贪心
- POJ 2976 浅谈二分答案+贪心
- POJ2018 Best Cow Fences——二分答案+贪心(动态规划)求最大子段和——pku2018
- POJ 3122 Pie(二分答案)
- POJ 3122 Pie【二分答案】
- poj3122——Pie(二分+贪心)
- 【POJ 3122】 Pie (二分+贪心)
- POJ 3122 Pie【二分答案】
- 【POJ 3122】 Pie (二分+贪心)
- POJ 3122 Pie 二分答案
- POJ 3122 Pie 二分答案
- [POJ 2976] Dropping tests 二分答案+伪贪心规划
- codeforces 349C Mafia [贪心]/[二分答案]
- 疫情控制 2012年NOIP全国联赛提高组(二分答案+贪心)
- [POJ]1743 不可重叠最长重复字串 二分答案
- poj 3294 求多于k个字符串的最长公共子串的个数-------后缀数组+二分答案
- poj 2456 最大化最小值(二分+贪心)