codeforces 987B. High School: Become Human

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本文介绍了一种快速比较xy与yx大小的方法，适用于特定条件下的整数输入，并提供了两种实现方式，一种采用对数运算，另一种则通过特殊值判断。

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Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.

It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.

One of the popular pranks on Vasya is to force him to compare $x^{y}$ with $y^{x}$ . Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.

Please help Vasya! Write a fast program to compare $x^{y}$ with $y^{x}$ for Vasya, maybe then other androids will respect him.

Input

On the only line of input there are two integers $x$ and $y$ ( $1 \leq x, y \leq 10^{9}$ ).

Output

If $x^{y} < y^{x}$ , then print '<' (without quotes). If $x^{y} > y^{x}$ , then print '>' (without quotes). If $x^{y} = y^{x}$ , then print '=' (without quotes).

Examples
inputCopy
5 8
outputCopy
>
inputCopy
10 3
outputCopy
<
inputCopy
6 6
outputCopy
=

Note

In the first example $5^{8} = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$ , and $8^{5} = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$ . So you should print '>'.

In the second example $10^{3} = 1000 < 3^{10} = 59049$ .

In the third example $6^{6} = 46656 = 6^{6}$ .

题意：给出x,y 问x

^{y} 和 y^{x}

的关系

分析：暴力的思路两边取log变成求ylogx和xlogy的关系，因为题目给出的是整数所以实际上没有想象中那么卡精度，等于那里再加个判断(x==y)

另外还有思路就是用证明的方法

xy>yx

可知越靠近e就越优秀，然后胡乱搞一波就行了

这里有个坑就是如果x=2,y=4时是等于

取log暴力判断的

#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
#include<set>
#include<cstdio>
#include<functional>
#include<iomanip>
#include<cmath>
#include<stack>
#include<iomanip>
#include<functional>
#include <iomanip>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int maxn = (int)1e5 + 100;
const int BN = 30;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;
const double eps = 1e-8;
const double PI = acos(-1);
int main() {
	double a, b;
	scanf("%lf%lf", &a, &b);
	double t1 = b*log(a), t2 = a*log(b);
	if (fabs(a - b) < eps || fabs(t1 - t2) < eps) printf("=\n");
	else if (t1 > t2) printf(">\n");
	else printf("<\n");
	return 0;
}

证明xjb搞的

#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
#include<set>
#include<cstdio>
#include<functional>
#include<iomanip>
#include<cmath>
#include<stack>
#include<iomanip>
#include<functional>
#include <iomanip>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int maxn = (int)1e5 + 100;
const int BN = 30;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;
const double eps = 1e-10;
const double PI = acos(-1);
int main() {
	LL a, b;
	while (~scanf("%I64d%I64d", &a, &b)) {
		if (a == b) printf("=\n");
		else if (a == 1) printf("<\n");
		else if (b == 1) printf(">\n");
		else if (a == 2) {
			if (b == 3) printf("<\n");
			else if (b == 4) printf("=\n");
			else printf(">\n");
		}
		else if (b == 2) {
			if (a == 3) printf(">\n");
			else if (a == 4) printf("=\n");
			else printf("<\n");
		}
		else {
			if (a > b) printf("<\n");
			else printf(">\n");
		}
	}
	return 0;
}

^{y} 和 y^{x}

^{y} 和 y^{x}