poj 3122 hdu 1969 Pie

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点击打开链接poj 3122

点击打开链接hdu' 1969

思路：二分
分析：
1 题目说的是有n块圆形的饼，饼的高度都是1但是半径不一定相同。现在有f+1个人（算上自己），想要平均分得一块面积相同的饼，必须是一块不能够是组合起来的，问这f+1个人能够分到的最大的面积是多少
2 很明显的二分ans，然后对n块饼进行判断，利用二分逼近的思想求出ans
3 注意这样一题的精度要求很高，pi = 3.1415926535898，还有eps不能取太小会TLE。

代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define MAXN 10010#define pi 3.1415926535898#define eps 1e-6int t , N , F;double sum;double r[MAXN];void solve(){   double left , right , mid;   left = 0;   right = 1.0*(pi*sum)/F;   while(right-left > eps){      mid = left+(right-left)/2.0;            int count = 0;      for(int i = 0 ; i < N ; i++){         double s = pi*r[i]*r[i];         int tmp = s/mid;         count += tmp;      }      if(count < F)         right = mid;      else         left = mid;   }   printf("%0.4lf\n" , right);}int main(){   scanf("%d" , &t);    while(t--){      scanf("%d%d" , &N , &F);      sum = 0;      F++;      for(int i = 0 ; i < N ; i++){          scanf("%lf" , &r[i]);          sum += r[i]*r[i];      }      solve();   }   return 0;}

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const double eps = 1e-6;const double PI = 3.1415926535898;const int MAXN = 10010;int n , numFriend;struct pie{    double r;    double size;};pie p[MAXN];bool judge(double s){     int cnt = 0;     for(int i = 0 ; i < n ; i++)         cnt += p[i].size/s;     return cnt >= numFriend+1 ? true : false;}double solve(double Min , double Max){     double left , right , mid;     left = Min , right = Max;     while(right-left > eps){         double mid = left+(right-left)/2;         if(judge(mid))            left = mid;         else            right = mid;     }     return left;}int main(){    int Case , pos;    double Min , Max;    scanf("%d" , &Case);    while(Case--){        scanf("%d%d" , &n , &numFriend);        Max = 0.0 , Min = 1>>30;        for(int i = 0 ; i < n ; i++){            scanf("%lf" , &p[i].r);            p[i].size = PI*p[i].r*p[i].r;            Max = max(Max , p[i].size);            Min = min(Min , p[i].size);        }        printf("%.4lf\n" , solve(Min , Max));     }    return 0;}