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自适应simpson积分公式

公式

通过二分区间递归求simpson积分

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=10+10,maxn=100000+10,inf=0x3f3f3f;

double v1,v2,k,x;
double f(double t)
{
    return k/(v1*v1*t*t+(x-v2*t)*(x-v2*t));
}
double simpson(double l,double r)
{
    return (f(l)+4*f((l+r)/2)+f(r))*(r-l)/6;
}
double solve(double l,double r)
{
    double m=(l+r)/2.0;
    double res=simpson(l,r);
    if(fabs(res-simpson(l,m)-simpson(m,r))<eps)return res;
    else return solve(l,m)+solve(m,r);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t;
    cin>>t;
    while(t--){
        cin>>v1>>v2>>x>>k;
        cout<<solve(0,1e18)<<endl;
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/acjiumeng/p/7126906.html