UVALive

题目：给你宽为W，深度为D的一块区域，再给你p1,q1,p2,q2；y1=p1/q1,y2=p2/q2，让你确定一个深度d,使得y1y2之间的面积在W*d这个区域内为A.

思路：二分+辛普森

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst double eps=1e-6;//根据实际情况修改double arr[4][10];double W,D,A,mid;int K;double F(double x,double *p,double *q){//Simpson公式用到的函数    double fz=0,fm=0;    for(int i=0;i<=K;i++){        fz=x*fz+p[i];        fm=x*fm+q[i];    }    return max(mid,fz/fm);}double simpson(double a, double b,double *p,double *q){double c = a + (b - a) / 2;return (F(a,p,q) + 4 * F(c,p,q) + F(b,p,q))*(b - a) / 6;}double asr(double a, double b, double eps, double A,double *p,double *q){double c = a + (b - a) / 2;double L = simpson(a, c,p,q), R = simpson(c, b,p,q);if (fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15.0;return asr(a, c, eps / 2, L,p,q) + asr(c, b, eps / 2, R,p,q);}double asr(double a, double b, double eps,double *p,double *q){return asr(a, b, eps, simpson(a, b,p,q),p,q);}int main(){    while(~scanf("%lf%lf%lf%d",&W,&D,&A,&K)){        for(int i=0;i<=3;i++)            for(int j=K;j>=0;j--)                scanf("%lf",&arr[i][j]);        double l=-D,r=0,ans;        while(r-l>eps){            mid=(l+r)/2;            double area=asr(0,W,eps,arr[0],arr[1])-asr(0,W,eps,arr[2],arr[3]);            if(area>=A){                ans=mid;                l=mid;            }            else r=mid;        }        printf("%.5f\n",f_abs(ans));    }    return 0;}