HDU

原题链接

题意：

给定 $a,b,l,r,$求 与$x = l,x = r$ 围成的封闭图形的面积。

思路：

大佬可以直接算一下原函数就出来了，当没法计算或者很难计算的时候就可以用 自适应$simpson$ 积分来逼近真实值。

/*
* @Author: windystreet
* @Date:   2018-08-04 16:24:01
* @Last Modified by:   windystreet
* @Last Modified time: 2018-08-04 16:24:33
*/

#include<bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define eps  1e-10                        // 精度根据题目改变
#define gcd __gcd
#define pb push_back
#define PI acos(-1.0)
#define lowbit(x) (x)&(-x)
#define bug printf("!!!!!
");
#define mem(x,y) memset(x,y,sizeof(x))

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef unsigned long long uLL;

const int maxn = 1e5+2;
const int INF  = 1<<30;
const int mod  = 1e9+7;

double a,b;

double f(double x){
    return b*(sqrt(1.0-(x*x)/(a*a)));    // 函数根据题目改变
}
double simpson(double L ,double R){
    return (R-L)*(f(L)+4.0*f((L+R)/2.0)+f(R))/6.0;
}
double asr(double L,double R){
    double mid = (L+R)/2.0;
    double res = simpson(L,R);
    double left = simpson(L,mid),right = simpson(mid,R);
    if(fabs(left+right-res)<eps)return left + right;
    else return asr(L,mid)+asr(mid,R);
}


void solve(){
    double l,r;
    scanf("%lf%lf%lf%lf",&a,&b,&l,&r);
    double ans = asr(l,r);
    printf("%.3lf
",2.0*ans);
    
    return;
}

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//    ios::sync_with_stdio(false);
    int t = 1;
    scanf("%d",&t);
    while(t--){
    //    printf("Case %d: ",cas++);
        solve();
    }
    return 0;
}