Codeforces621DRatKweshandCheese

传送门：
http://codeforces.com/contest/621/problem/D

double 1.07*10的308位
long double 简直神器 4900多位
此题log里面会卡负数，因此可以用long double 然后一个log 就可以了！！

今天想叉掉一个B题暴力写的，然后输出文件手点的太快，导致数据没写完，最后invalid input 简直可惜啊！！！！

#include 
#include 

#define double long double
using namespace std;

double get2(double x, double y, double z) {
    //(x ^ y) ^ z
    return y * z * log(x);
}

const double eps = 1e-10;

bool bigger(double a, double b) {
    //a > b
    return a - b > 0; //-eps;
}

bool equal(double a, double b) {
    return abs(a - b) double best;
int best_pos;
int pos;

void try1(double x, double y, double z, bool sure = false) {
    ++ pos;

    //x ^ y ^ z
    double val = pow(y, z) * log(x);

    if (bigger(val, best) || sure) {
        best = val;
        best_pos = pos;
    }
}

void try2(double x, double y, double z, bool sure = false) {
    ++ pos;

    //(x ^ y) ^ z
    double val = y * z * log(x);

    if (bigger(val, best) || sure) {
        best = val;
        best_pos = pos;
    }
}


int main()
{
    double x, y, z;
    cin >> x >> y >> z;

    try1(x, y, z, true);
    try1(x, z, y);
    try2(x, y, z);
    try2(x, z, y);

    try1(y, x, z);
    try1(y, z, x);
    try2(y, x, z);
    try2(y, z, x);

    try1(z, x, y);
    try1(z, y, x);
    try2(z, x, y);
    try2(z, y, x);
    //cout<
    if (best_pos == 1)
        cout <<"x^y^z\n";
    else if (best_pos == 2)
        cout <<"x^z^y\n";
    else if (best_pos == 3)
        cout <<"(x^y)^z\n";
    else if (best_pos == 4)
        cout <<"(x^z)^y\n";
    else if (best_pos == 5)
        cout <<"y^x^z\n";
    else if (best_pos == 6)
        cout <<"y^z^x\n";
    else if (best_pos == 7)
        cout <<"(y^x)^z\n";
    else if (best_pos == 8)
        cout <<"(y^z)^x\n";
    else if (best_pos == 9)
        cout <<"z^x^y\n";
    else if (best_pos == 10)
        cout <<"z^y^x\n";
    else if (best_pos == 11)
        cout <<"(z^x)^y\n";
    else if (best_pos == 12)
        cout <<"(z^y)^x\n";

    return 0;
}

不过题解给的思想还是蛮给力的，毕竟long double
能解决的位数也是有限的，属于是水过，标解是这样的，我们还是取双log，分成三类，
1.如果a,b,c都是大于1的，那么好办
2.如果至少有一个是大于1的，那么就要把大于1的作为底，因为如果把小于1的做底，那么结果永远是小于1的，那么我们就可以不去比较那些会使得值为负数的表达式，就ok了
3.如果三个数都小于1的话，那么我们可以将底数取倒数，然后就比较最小值就ok了，简直机智啊！orz