求单峰函数的最值点

    1. #include <bits/stdc++.h>
    2. using namespace std;
    3. const int N = 15;
    4. const double eps = 1e-12;
    5. int n;
    6. double a[N];
    7. double cal(double x) {
    8.     double pow = 1, res = 0;
    9.     for (int i = n+1; i >= 1; i--) {
    10.         res += a[i]*pow;
    11.         pow *= x;
    12.     }
    13.     return res;
    14. }    
    15. signed main() {
    16.     double l, r;
    17.     cin >> n >> l >> r;
    18.     for (int i = 1; i <= n+1; i++)
    19.         scanf("%lf", &a[i]);
    20.     while (r-l >= eps) {
    21.         double m1 = l+(r-l)/3;
    22.         double m2 = r-(r-l)/3;
    23.         //此处求的是最大值，最小值改为<=即可
    24.         if (cal(m2)-cal(m1) >= eps) {
    25.             l = m1;
    26.         } else {
    27.             r = m2;
    28.         }
    29.     }
    30.     printf("%.5lf\n", l);
    31.     return 0;
    32. }