HDU 1969 Pie

来源：互联网发布：Java中怎样输出Path 编辑：程序博客网时间：2024/05/29 15:55

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 10005int 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;}

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