Codeforces Round #341 Div.2 D. Rat Kwesh and Cheese

嗯本来想着直接算出来不就行了吗 然后我想到了200^200^200.......

好吧其实也不难取两次log就行了 然后我第一次写出来log就写残了...........

log里面的拆分要仔细啊...

果然涉及到log就容易出问题 一组小于等于1的数据直接就跪了= = 

不过我猜跪在这组数据的人应该蛮多的....

解决办法的话应该是对这个数据不log处理吧...懒得写了= =

#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define INF 0x3f3f3f3f
#define mem(str,x) memset(str,(x),sizeof(str))  
#define STOP puts("Pause");
using namespace std;
typedef long long LL;

int main()
{
    double x, y, z;
    double data[20];
    scanf("%lf%lf%lf", &x, &y, &z);
    data[1] = z * log(y) + log(log(x));
    data[2] = y * log(z) + log(log(x));
    data[3] = log(y * z * log(x));
    data[4] = data[3];
    data[5] = z * log(x) + log(log(y));
    data[6] = x * log(z) + log(log(y));
    data[7] = log(z * x * log(y));
    data[8] = data[7];
    data[9] = y * log(x) + log(log(z));
    data[10] = x * log(y) + log(log(z));
    data[11] = log(y * x * log(z));
    data[12] = data[11];
    int ans = 1;
    for(int i = 2; i <= 12; i++){
        if(data[i] > data[ans]) ans = i;
    }
    switch(ans){
        case 1:puts("x^y^z");break;
        case 2:puts("x^z^y");break;
        case 3:puts("(x^y)^z");break;
        case 5:puts("y^x^z");break;
        case 6:puts("y^z^x");break;
        case 7:puts("(y^x)^z");break;
        case 9:puts("z^x^y");break;
        case 10:puts("z^y^x");break;
        case 11:puts("(z^x)^y");break;
    }
    return 0;
}