原題鏈接

題意：

給定一個邊長為n的正方形，求陰影部分面積

思路：

現將圖形順時針旋轉 45° 然後建立坐標系，寫出陰影部分方程，用Simpson積分算一下就行了，註意精度，1e-10 WA 了 ，1e-20 A了

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define X first
 6 #define Y second
 7 #define eps  1e-20
 8 #define gcd __gcd
 9 #define pb push_back
10
 
 #define PI acos(-1.0)
11 #define lowbit(x) (x)&(-x)
12 #define bug printf("!!!!!\n");
13 #define mem(x,y) memset(x,y,sizeof(x))
14 
15 typedef long long LL;
16 typedef long double LD;
17 typedef pair<int,int> pii;
18 typedef unsigned long long uLL;
19 
20 const int maxn = 1e5+2;
 
21 const int INF  = 1<<30;
22 const int mod  = 1e9+7;
23 
24 double f(double x){
25     return  sqrt(1-x*x) - sqrt(4 - x*x) + sqrt(2);
26 }
27 double simpson(double L ,double R){
28     return (R-L)*(f(L)+4.0*f((L+R)/2.0)+f(R))/6.0;
29 }
30 double asr(double L,double R){
31     double mid = (L+R)/2.0
 
;
32     double res = simpson(L,R);
33     double left = simpson(L,mid),right = simpson(mid,R);
34     if(fabs(left+right-res)<eps)return left + right;
35     else return asr(L,mid)+asr(mid,R);
36 }
37 double C;
38 void solve(){
39     int n;
40     scanf("%d",&n);
41     printf("%.2f\n",n*n*1.0*C);
42     return;
43 }
44 
45 int main()
46 {
47 //    freopen("in.txt","r",stdin);
48 //    freopen("out.txt","w",stdout);
49 //    ios::sync_with_stdio(false);
50     int t = 1;
51     scanf("%d",&t);
52     C = asr(0,sqrt(7.0/8.0));       // 離線計算單位系數
53     while(t--){
54     //    printf("Case %d: ",cas++);
55         solve();
56     }
57     return 0;
58 }

HDU - 5858 Hard Problem (simpson積分)