【BZOJ2882】工艺【最小表示法】

#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 600005;

int n, s[maxn];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

int MR() {
	int i = 0, j = 1;
	for(int k; i < n && j < n; ) {
		for(k = 0; k < n && s[i + k] == s[j + k]; k++);
		if(k == n) return i;
		if(s[i + k] > s[j + k]) i = i + k + 1;
		else j = j + k + 1;
		if(i == j) j++;
	}
	return min(i, j);
}

int main() {
	n = iread();
	for(int i = 0; i < n; i++) s[n + i] = s[i] = iread();

	int pos = MR();

	for(int i = 0; i < n - 1; i++) printf("%d ", s[pos + i]); printf("%d\n", s[pos + n - 1]);

	return 0;
}