牛客網-Infinite String Comparision

志のある者は事が成し遂げられ、釜を破って舟を沈め、百二秦関は結局楚に属した.苦心している人は天に負けず、臥薪嘗胆、三千越甲は呉を飲み込むことができる!
Infinite String Comparision
タイトルの説明
For a string xx, Bobo defines x∞ = =xxx…, which is xx repeats for infinite times, resulting in a string of infinite length. Bobo has two strings aa and bb. Find out the result comparing a∞ and b∞ in lexicographical order. You can refer the wiki page for further information of Lexicographical Order.
入力
The input consists of several test cases terminated by end-of-file. The first line of each test case contains a string a, and the second line contains a string b.

1 ≤|a|, |b| ≤ 105

a, b consists of lower case letters.

The total length of input strings does not exceed 2*106

しゅつりょく
For each test case

print “= ” (without quotes) if a∞ =b∞

Otherwise, print “< ” if a∞ < b∞

or “> ” if a∞> b∞

Sample Input

aa
b
zzz
zz
aba
abaa

Sample Output

<
=
>

問題を解く構想の無限循環はずっと循環する必要はなく、無限循環が持つべき性質を体現させるだけでよい.関連先に穴があるにすぎない.最大長の文字列を2倍に広げます.これにより、すべての詳細がこの新しい文字列に反映されます.次に、同じ長さになるまで小さな文字列を拡張します.このとき、2つの新しい文字列がこの無限ループの文字列を表すことができます.
AC時間～

#include
#include
#include
#include 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#pragma warning(disable:4244)
#define PI 3.141592653589793
#pragma GCC optimize(2)
#define accelerate cin.tie(NULL);cout.tie(NULL);ios::sync_with_stdio(false);
#define EPS 1.0e-8
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll ll_inf = 9223372036854775807;
const int int_inf = 2147483647;
const short short_inf = 32767;
const char char_inf = 127;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll read() {
	ll c = getchar(), Nig = 1, x = 0;
	while (!isdigit(c) && c != '-')c = getchar();
	if (c == '-')Nig = -1, c = getchar();
	while (isdigit(c))x = ((x << 1) + (x << 3)) + (c ^ '0'), c = getchar();
	return Nig * x;
}
inline void out(ll a) {
	if (a < 0)putchar('-'), a = -a;
	if (a >= 10)out(a / 10);
	putchar(a % 10 + '0');
}
ll phi(ll n)
{
	ll ans = n, mark = n;
	for (ll i = 2; i * i <= mark; i++)
		if (n % i == 0) { ans = ans * (i - 1) / i; while (n % i == 0)n /= i; }
	if (n > 1)ans = ans * (n - 1) / n; return ans;
}
ll qpow(ll x, ll n, ll mod) {
	ll res = 1;
	while (n > 0) {
		if (n & 1)res = (res * x) % mod;
		x = (x * x) % mod;
		n >>= 1;
	}
	return res;
}
ll mat_mod;
struct Mat {
	ll m[10][10];
};
Mat Mul(Mat A, Mat B, ll mat_size) {
	Mat res;
	memset(res.m, 0, sizeof(res.m));
	for (int i = 0; i < mat_size; i++)for (int j = 0; j < mat_size; j++)for (int k = 0; k < mat_size; k++)
		res.m[i][j] = (res.m[i][j] + (A.m[i][k] * B.m[k][j]) % mat_mod) % mat_mod;
	return res;
}
Mat mat_qpow(Mat data, ll power, ll mat_size) {
	Mat res;
	memset(res.m, 0, sizeof(res.m));
	for (int i = 0; i < mat_size; i++)res.m[i][i] = 1;
	while (power) {
		if (power & 1)res = Mul(res, data, mat_size);
		data = Mul(data, data, mat_size), power >>= 1;
	}
	return res;
}
#define Floyd for(int k = 1; k <= n; k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)
#define read read()
string c(string b, int n, int l)
{
	string res = b;
	for (int i = 0; i < n; i++)res += b;
	return res.substr(0, l);
}
int main()
{
	string a, b;
	while (cin >> a >> b)
	{
		if (a.size() >= b.size())
		{
			a = a + a;
			int la = a.size();
			int lb = b.size();
			int k = la / lb;
			k++;
			b = c(b, k, la);
			if (a > b)cout << ">" << endl;
			else if (a < b)cout << " << endl;
			else cout << "=" << endl;
		}
		else
		{
			swap(a, b);
			a = a + a;
			int la = a.size();
			int lb = b.size();
			int k = la / lb;
			k++;
			b = c(b, k, la);
			if (a > b)cout << " << endl;
			else if (a < b)cout << ">" << endl;
			else cout << "=" << endl;
		}
	}
}

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