UVALive - 3635 - Pie（二分）

来源：互联网发布：苹果在线软件下载编辑：程序博客网时间：2024/04/30 01:56

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 << endltypedef 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;}

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