题解 | 解方程

#include <bits/stdc++.h>//这是我这个小白写过最ex的二分题
using namespace std;
using ld = long double;
using ll = long long;

ll a, b, c;
ld eps = 1e-6;  

bool check(ld x) {
    ld res = pow(x, (ld)a) + (ld)b * log(x);
    return res >= c; 
}

int main() {
    cin >> a >> b >> c;
 
    ld res_at_1 = pow(1.0, (ld)a) + (ld)b * log(1.0);
    if (fabs(res_at_1 - c) <= 1e-7) {
        cout << fixed << setprecision(12) << 1.0 << endl;
        return 0;
    }
    
    ld l = 1.0;
    ld r = 1.0;
     
    if (res_at_1 < c) {
  
        while (pow(r, (ld)a) + (ld)b * log(r) < c) {
            r *= 2.0;
        }
    } else {
  
        cout << "-1" << endl;
        return 0;
    }
     
    for (int i = 0; i < 100; i++) {
        ld mid = (l + r) / 2.0;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid;
        }
    }
   
    ld x = r;
    ld res = pow(x, (ld)a) + (ld)b * log(x);
  
    if (fabs(res - c) <= 1e-7) {
        cout << fixed << setprecision(12) << x << endl;
    } else {
        cout << "-1" << endl;
    }
    
    return 0;
}