poj 3122 Pie (二分)

题意：给出若干个pie的半径，从这些pie中切出m个大小相等形状任意的块。可以剩余边角余料并扔掉。问每块最大多大。

分析：这题是浮点数的二分题，二分每块的大小。

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

double r[10010];
int n, f;

bool C ( double x )
{
    int num = 0;
    for ( int i = 0; i < n; i++ )
        num += ( int ) ( r[i] / x );
    return num >= f;
}
int main()
{
    int t;
    scanf ( "%d", &t );
    while ( t-- )
    {
        scanf ( "%d%d", &n, &f );
        f++;
        for ( int i = 0; i < n; i++ )
        {
            scanf ( "%lf", &r[i] );
            r[i] = PI * r[i] * r[i];
        }
        double lb = 0, ub = 100000000 * PI;
        for ( int i = 0; i < 100; i++ )
        {
            double mid = ( lb + ub ) / 2;
            if ( C ( mid ) )
                lb = mid;
            else
                ub = mid;
        }
        printf ( "%.4f\n", ub );
    }
    return 0;
}