UVA 1356 Bridge 自适应辛普森

题意：桥上等距离放着若干的塔，高为H，相邻两座塔之间的距离不得超过D，塔之间有绳
索，为抛物线，桥长度为B，绳索长度为L，问你建造最少塔时绳索最下端距离地面的高度。

思路：注意到相同的宽度下，抛物线高度和长度成正比，所以我们二分高度，求长度，抛物线的积分公式为√（1+(f’(x))^2）,利用辛普森公式就能很轻松的求出~

第一次用辛普森，好激动~

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
using namespace std;

const double eps = 1e-8;

double D, H, B, L;
double a;
int cas = 0;

double F(double x){
    return sqrt(1 + 4 * a * a * x * x);
}

double simpson(double a, double b){
    double c = a + (b - a) / 2;
    return (F(a) + 4 * F(c) + F(b)) * (b - a) / 6;
}

double asr(double a, double b, double eps, double A){
    double c = a + (b - a) / 2;
    double L = simpson(a, c), R = simpson(c, b);
    if(fabs(L + R - A) <= eps) return L + R + (L + R - A) / 15;
    return asr(a, c, eps / 2, L) + asr(c, b, eps / 2, R);
}

double asr(double a, double b, double eps){
    return asr(a, b, eps, simpson(a, b));
}

void solve(){
    scanf("%lf%lf%lf%lf", &D, &H, &B, &L);
    int n = ceil(B / D);
    double D1 = B / n;
    double L1 = L / n;
    double l = 0, r = H;
    while(l + eps < r){
        double mid = (l + r) / 2;
        a = 4 * mid / (D1 * D1);
        if(2 * asr(0, D1 / 2, 1e-8) > L1) r = mid;
        else l = mid;
    }
    if(cas++) printf("\n");
    printf("Case %d:\n", cas);
    printf("%.2lf\n", H - l);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--) solve();
    return 0;
}