poj_3122

最新推荐文章于 2019-07-21 21:58:23 发布

ThinkingLion

最新推荐文章于 2019-07-21 21:58:23 发布

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分类专栏：数学文章标签：二分法

本文链接：https://blog.csdn.net/ThinkingLion/article/details/44854293

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源地址：http://poj.org/problem?id=3122

题目大意就是有n块馅饼，要平均分成f份，问最大每份是多少。

二分即可，上限是所有馅饼的面积和，下限是0.

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-5
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
#define pi acos(-1.0);
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
double area[10010];
int f;
bool Check(double Area){
	int num=0;
	for(int i=0;i<n;i++){	//calculate when the area is Area,how many pies can be divided
		double tmp = area[i];
		while(tmp >= Area){
			tmp -= Area;
			num++;
		}
	}
	return (num>f);
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    int t;
    scanf("%d",&t);
    while(t--){
    	scanf("%d%d",&n,&f);
    	int x;
    	double r = 0;
    	for(int i=0;i<n;i++){
    		scanf("%d",&x);
    		area[i] = x*x*pi;
    		r += area[i];
    	}
    	double l = 0.0;
    	while(r-l>=eps){
    		double mid = (r+l)/2;
    		if(Check(mid))
    			l = mid;
    		else
    			r = mid;
    	}
    	printf("%.4f\n",l);
    }
return 0;
}