LA 3485 辛普森公式求积分

传送门 : LA 3485

题解

1. 曲线积分知识（弧长公式）：
   若可导曲线函数为f(x)其弧长公式为
   L=∫ba1+f′(x)2−−−−−−−−√
   所以这题抛物线设为f(x) = a(x - d)(x +d)
   弧长为
   2∗∫d01+4a2x2−−−−−−−−√
2. 先预处理出d,可以看出ｄ固定时弧长随ａ递增而递增，　所以直接二分枚举ａ＊ｄ＊ｄ求解，　积分用辛普森公式．

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code

#include <iostream>#include <iomanip>#include <cmath>using namespace std;typedef double ld;const ld eps = 1e-10;int D, H, B, L;ld a;ld f(ld x){ return sqrt(1 + 4 * a * a * x * x);}ld cal(ld l, ld r){    return (r - l) * (f(l) + 4 * f((l + r) / 2) + f(r)) / 6;}ld simpson(ld l, ld r){    ld mid = (l + r) / 2;    ld sa = cal(l, r);    ld la = cal(l, mid), ra = cal(mid, r);    if(fabs(sa - la - ra) < eps) return sa;    return simpson(l, mid) + simpson(mid, r);}int main(){    int t, cas = 0;    cin >> t;    while(t--){        cin >> D >> H >> B >> L;        int n = (B + D - 1) / D;/**不加D-1,RE...*/        ld ll = (ld)L / n, dd = (ld)B / n / 2;/**预处理dd*/        ld l = 0, r = (ld)H;        while(r - l > eps){/**二分*/            ld mid = (l + r) / 2;            a = mid / dd / dd;/**(mid = a * d * d) */            if(simpson(0, dd) * 2 > ll) r = mid;            else l = mid;        }        cout << "Case " << ++cas << ":" << endl;        cout << fixed << setprecision(2) << H - l << endl;        if(t) cout << endl;    }    return 0;}