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  • HDU 4498 Function Curve (分段, simpson)

    转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove

    最近太逗了。。。感觉成都要打铁了。。。只能给队友端茶送水了。。。。

    积分都不会了。。。曲线长度不会求。。。。

    写个代码,一堆SB错误。。。。。

    纯属吐槽博文 。。。。。。

    解法 :首先把n个函数以及y = 100求出交点。。。。把交点排序。

    然后 处理每个区间,求出这段要积的函数

    由于sqrt (1 + x ^ 2)不会求不定积分。。。只能simpson一下了。。。

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <vector>
    #include <algorithm>
    using namespace std;
    const int N = 55;
    const double eps = 1e-10;
    int n;
    double a[N] , b[N] , k[N];
    vector <double> inter;
    int dcmp (double d) {
        return d < -eps ? -1 : d > eps;
    }
    double sqr (double d) {
        return d * d;
    }
    void check (double d) {
        if (dcmp (d) >= 0 && dcmp (d - 100) <= 0)
            inter.push_back (d);
    }
    void get_inter () {
        for (int i = 0 ; i < n ; i ++) {
            if (dcmp (b[i] - 100) > 0) continue;
            double x1 = sqrt ((100 - b[i]) / k[i]) + a[i];
            double x2 = -sqrt ((100 - b[i]) / k[i]) + a[i];
            check (x1) ; check (x2);
        }
        for (int i = 0 ; i < n ; i ++) {
            for (int j = i + 1 ; j < n ; j ++) {
                double A = (k[i] - k[j]);
                double B = -(2 * k[i] * a[i] - 2 * k[j] * a[j]);
                double C = k[i] * a[i] * a[i] + b[i] - k[j] * a[j] * a[j] - b[j];
                if (dcmp (A) == 0) {
                    if (dcmp (B)) check (-C / B);
                    continue;
                }
                if (B * B - 4 * A * C < 0) continue;
                if (dcmp (B * B - 4 * A * C) == 0) check (-B / 2 / A);
                else {
                    double delta = sqrt (B * B - 4 * A * C);
                    double x1 = (-B + delta) / 2 / A , x2 = (-B - delta) / 2 / A;
                    check (x1); check (x2);
                }
            }
        }
    }
    
    double Function (double x , int i) {
        return k[i] * sqr (x - a[i]) + b[i];
    }
    int best;
    double function (double x) {
        return sqrt (1 + sqr (2 * k[best] * (x - a[best])));
    }
    double simpson (double l , double r ) {  
        return (function (l ) + 4 * function ((l + r) / 2.0 ) + function (r )) * (r - l) / 6.0;  
    }  
    double simpson (double l , double r , double all , double eps) {
        double m = (l + r) / 2.0;
        double L = simpson (l , m) , R = simpson (m , r);
        if (fabs (L + R - all) <= 15 * eps) return L + R + (L + R - all) / 15;
        return simpson (l , m , L , eps / 2.0) + simpson (m , r , R , eps / 2.0);
    }
    double simpson (double l , double r , double eps) {
        return simpson (l , r , simpson (l , r) , eps);
    }
    int main () {
        #ifndef ONLINE_JUDGE
            freopen ("input.txt" , "r" , stdin);
            // freopen ("output.txt" , "w" , stdout);
        #endif
        int t ;
        scanf ("%d" , &t);
        while (t --) {
            inter.clear ();
            scanf ("%d" , &n);
            for (int i = 0 ; i < n ; i ++) {
                scanf ("%lf %lf %lf" , &k[i] , &a[i] , &b[i]);
            }
            get_inter ();
            inter.push_back (0); inter.push_back (100);
            sort (inter.begin () , inter.end ());
            int size = inter.size() ;
            double ans = 0;
            for (int i = 1 ; i < size ; i ++) {
                double x1 = inter[i - 1] , x2 = inter[i];
                if (dcmp (x1 - x2) >= 0) continue;
                double m = (x1 + x2) / 2.0;
                best = 0;
                for (int j = 1 ; j < n ; j ++) {
                    if (dcmp (Function (m , j) - Function (m , best)) < 0)
                        best = j;
                }
                if (dcmp (Function (m , best) - 100) >= 0) {
                    ans += (x2 - x1);
                    continue;
                }
                ans += simpson (x1 , x2  , 1e-8);
            }
            printf ("%.2f
    " , ans);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/riasky/p/3360884.html
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