HDU 1969 Pie

最新推荐文章于 2019-05-09 23:24:20 发布

原创最新推荐文章于 2019-05-09 23:24:20 发布 · 4.3k 阅读

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分蛋糕问题要求每一块蛋糕体积相等其实就是求r[i]能分解为f+1个相同的数的最大值二分水过

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<cmath>
#include<list>
#include<cstring>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>

#define sci(a) scanf("%d",&a)
#define scd(a) scanf("%lf",&a)
#define scs(a) scanf("%s",a)
#define pri(a) printf("%d\n",a);
#define prd4(a) printf("%.4lf\n",a);
#define prs(a) printf("%s\n",a);

const double eps=1e-6;
#define N 10005
int r[N];//半径
const double PI=4.0*atan(1.0);

bool check(double size,int n,int f){//能否分出f个size大小的蛋糕
    int sum=0;
    for(int i=0;i<n;++i){
        sum+=(int)(r[i]/size);//这里好奇怪 如果加上eps居然WA
        if(sum>=f)
            return true;
        if(r[i]<=size+eps)
            return false;
    }
    return false;
}

double Bin_Search(int n,int f){//二分找最大size
    double MAX=r[0],MIN=0,mid,res=0;
    while(MAX-MIN>eps){
        mid=(MIN+MAX)/2;
        if(check(mid,n,f)){
            MIN=mid;
            res=mid;
        }
        else
            MAX=mid;
    }
    return res;
}

int main(){
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);

    int t,n,f;
    sci(t);
    while(t--){
        sci(n);
        sci(f);
        ++f;
        for(int i=0;i<n;++i){
            sci(r[i]);
            r[i]*=r[i];
        }
        sort(r,r+n,greater<int>());
        prd4(Bin_Search(n,f)*PI)
    }
    return 0;
}