Codeforces 621 D Rat Kwesh and Cheese

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 <iostream>#include <cmath>#define double long doubleusing 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) < eps;}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<<best<<endl;    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