题目链接：传送门

题意就不说了，直接说怎么推的

formula

formula

直接在     求的结果前面+2个log 取对数

loglog(f(a)^f(b)^f(c)) = (log(f(b)^f(c))*log(f(a)) = log(f(b)^f(c)) + loglog(f(a)) = f(c)log(f(b)) + loglog(f(a)) = f(c)f(b+1) + f(a+2)

f(+无穷大) 趋近于1

f(n)   n越大 f（n）越小

f(a) 与 f(b)f(c)比较  可化为  f(a)f(+无穷大) 与 f(b)f(c)比较

a+b 与 c+d进行比较 比较max(a,b) 与 max(c,d),相同则比较min(a,b) 与 min(c,d)
AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define inf 0x3f3f3f3f

using namespace std;

struct node {
	int x, y;
	node(int _x, int _y){//x 为小值   y 为大值 
		x = _x;
		y = _y;
		x = min(x, inf);
		y = min(y, inf);
		if(x > y){
			swap(x, y);
		}
	}
};

int a[3];
int b[3];

int compare(node aa, node bb){
	if(aa.x < bb.x){
		return 1;
	}else if(aa.x > bb.x){
		return -1;
	}
	if(aa.y < bb.y){
		return 1;
	}else if(aa.y > bb.y){
		return -1;
	}
	return 0;
} 
int check(node aa, node bb, node cc, node dd) {
	if(compare(aa, cc)){
		return compare(aa, cc);
	}else {
		return compare(bb, dd);
	}
}
int main() {
	int t;
	scanf("%d", &t);
	while(t--){
		memset(a, inf , sizeof(a));
		memset(b, inf , sizeof(b));
		int n, m;
		scanf("%d%d", &n, &m);
		for(int i = 0; i < n; i++){
			scanf("%d", &a[i]);
		}
		for(int i = 0; i < m; i++){
			scanf("%d", &b[i]);
		}
		node k1 = node(a[0]+2, inf), k2 = node(a[1]+1, a[2]);
		node k3 = node(b[0]+2, inf), k4 = node(b[1]+1, b[2]);
		if(compare(k1, k2) == -1){
			swap(k1, k2);// k1 > k2
		}
		if(compare(k3, k4) == -1){
			swap(k3, k4);// k3 > k4
		}
		printf("%d\n",check(k1,k2,k3,k4));
		
	}
	return 0;
}