UVALive - 3635 - Pie（二分）

Posted 2020-08-15 TianTengtt

题意：有F + 1（1 <= F <= 10000）个人分N（1 <= N <= 10000）个圆形派，每个人得到的派面积相同，且必须是一整块（不能够两个甚至多个派拼在一起），求每个人最多能得到多大面积的派。（误差最多到0.001）

因为答案是小数类型的，并且N高达10000，故不可暴力枚举。

可以二分枚举最大面积，然后检查是否切出来派的总个数大于等于F + 1。

（判相等时不可直接判相等，需要加精度控制）

 

#include<cstdio>  
#include<cstring>  
#include<cctype>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<deque>  
#include<queue>  
#include<stack>  
#include<list>  
#define fin freopen("in.txt", "r", stdin)  
#define fout freopen("out.txt", "w", stdout)  
#define pr(x) cout << #x << " : " << x << "   "  
#define prln(x) cout << #x << " : " << x << endl  
typedef long long ll;  
typedef unsigned long long llu;  
const int INT_INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;  
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const double pi = acos(-1.0);  
const double EPS = 1e-6;  
const int dx[] = {0, 0, -1, 1};  
const int dy[] = {-1, 1, 0, 0};  
const ll MOD = 1e9 + 7;  
const int MAXN = 100 + 10;  
const int MAXT = 10000 + 10;  
using namespace std;  
  
int T, n, f;  
double a[MAXT];  
  
bool judge(double area){  
    int sum = 0;  
    for(int i = 0; i < n; ++i)  sum += int(a[i] / area);  
    return sum >= f;  
}  
  
int main(){  
    scanf("%d", &T);  
    while(T--){  
        scanf("%d%d", &n, &f);  
        for(int i = 0; i < n; ++i){  
            scanf("%lf", a + i);  
            a[i] = a[i] * a[i] * pi;  
        }  
        ++f;  
        double l = 0.0, r = *max_element(a, a + n);  
        while(l + EPS < r){  
            double mid = (l + r) / 2;  
            if(judge(mid))  l = mid;  
            else  r = mid;  
        }  
        printf("%.4lf\n", l);  
    }  
    return 0;  
}