UVaLive 3635 Pie (二分)

题意：有f+1个人来分n个圆形派，每个人得到的必须是一个整块，并且是面积一样，问你面积是多少。

析：二分这个面积即可，小了就多余了，多了就不够分，很简单就能判断。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];

bool judge(double mid){
    int ans = 0;
    for(int i = 0; i < n; ++i)
        ans += (int)floor(PI * a[i] * a[i] / mid);
    return ans >= m;
}

double solve(){
    double l = 0, r = PI * 10000.0* 10000.0;
    while(fabs(r - l) > 1e-5){
        double mid = (l + r) / 2.0;
        if(judge(mid))  l = mid;
        else r = mid;
    }
    return l;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        ++m;
        for(int i = 0; i < n; ++i)  scanf("%d", a+i);
        printf("%.4f\n", solve());
    }
    return 0;
}