codeforce341 D題 Rat Kwesh and Cheese（複數）

題意

計算出由x,y,z構成的下標最小的 值最大的表達式。
0.1 ≤ x, y, z ≤ 200.0

思路

因爲xyz 可能非常大,所以需要使用2次log函數。可以將xyz 轉換成z∗log(y)+log(log(x)) 但是如果x小於1的時候log(x) 是小於0的。 所以參考別人的思路後，覺得是可以使用複數來解決。 任何一個複數a+b∗i 的對數都可以表示成lnr+i∗θ 。對於2個負數，如果值越大，則r越小，則其log值得實部越小。

代碼

（原大牛作者版本http://codeforces.com/profile/thuustalu）

#include <iostream>
#include <complex>
#include <string>

using namespace std;

bool bigger (complex<long double> a, complex<long double> b) {
  if (imag(a) == 0 && imag(b) == 0) {
    return real(a) > real(b);
  } else if (imag(a) == 0 && imag(b) != 0) {
    return true;
  } else if (imag(a) != 0 && imag(b) == 0) {
    return false;
  } else if (imag(a) != 0 && imag(b) != 0) {
    return real(a) < real(b);
  }
}

int main () {
  long double ax, ay, az;
  cin >> ax >> ay >> az;

  complex<long double> x (ax, 0.0L);
  complex<long double> y (ay, 0.0L);
  complex<long double> z (az, 0.0L);

  complex<long double> cmaz (3, 3);
  string ans = "xd";

  if (bigger(z * log(y) + log(log(x)), cmaz)) {
    cmaz = z * log(y) + log(log(x));
    ans = "x^y^z";
  }
  if (bigger(y * log(z) + log(log(x)), cmaz)) {
    cmaz = y * log(z) + log(log(x));
    ans = "x^z^y";
  }
  if (bigger(log(y * z) + log(log(x)), cmaz)) {
    cmaz = log(y * z) + log(log(x));
    ans = "(x^y)^z";
  }

  if (bigger(z * log(x) + log(log(y)), cmaz)) {
    cmaz = z * log(x) + log(log(y));
    ans = "y^x^z";
  }
  if (bigger(x * log(z) + log(log(y)), cmaz)) {
    cmaz = x * log(z) + log(log(y));
    ans = "y^z^x";
  }
  if (bigger(log(x * z) + log(log(y)), cmaz)) {
    cmaz = log(x * z) + log(log(y));
    ans = "(y^x)^z";
  }

  if (bigger(y * log(x) + log(log(z)), cmaz)) {
    cmaz = y * log(x) + log(log(z));
    ans = "z^x^y";
  }
  if (bigger(x * log(y) + log(log(z)), cmaz)) {
    cmaz = x * log(y) + log(log(z));
    ans = "z^y^x";
  }
  if (bigger(log(x * y) + log(log(z)), cmaz)) {
    cmaz = log(x * y) + log(log(z));
    ans = "(z^x)^y";
  }

  cout << ans << endl;
}