HDU 2609 How many

https://cn.vjudge.net/problem/HDU-2609

题目

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me
How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some).
For example 0110 express a necklace, you can rotate it. 0110 -> 1100 -> 1001 -> 0011->0110.

Input

The input contains multiple test cases.

Each test case include: first one integers n. (2<=n<=10000)
Next n lines follow. Each line has a equal length character string. (string only include '0','1').

Output

For each test case output a integer , how many different necklaces.

Sample Input

4
0110
1100
1001
0011
4
1010
0101
1000
0001

Sample Output

1
2

题解

$mathcal{O}(n)$贪心最小表示法，然后排序= =

虽然可以Hash+EXKMP判断，但是计算HASH时也需要最小表示法……

AC代码

#include<bits/stdc++.h>
#define REP(r,x,y) for(register int r=(x); r<(y); r++)
#define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
#define PERE(r,x,y) for(register int r=(x); r>=(y); r--)
#ifdef sahdsg
#define DBG(...) printf(__VA_ARGS__), fflush(stdout)
#else
#define DBG(...) (void)0
#endif // sahdsg
using namespace std;
char buf[207];
inline void getmin(char *p, int l) {
	char *a=p, *b=p+1;
	while(a-p<l && b-p<l) {
		int i=0;
		while(i<l && a[i]==b[i]) i++;
		if(i==l) break;
		if(a[i]<b[i]) {
			b+=i+1;
		} else if(a[i]>b[i]) {
			a+=i+1;
		}
		if(a==b) b++;
	}
	if(a-b>0) a=b;
	memcpy(buf,a,l);
	memcpy(p,buf,l); p[l]=0;
}
char s[10007][207];
int z[10007];
int l;
inline bool cmp(int a, int b) {
	return strcmp(s[a],s[b])<0;
}
inline bool eq(int a, int b) {
	return strcmp(s[a],s[b])==0;
}
int main() {
	int n;
	while(~scanf("%d", &n)) {
		REP(i,0,n) {
			scanf("%s", s[i]);
			z[i]=i;
			if(!i) l=strlen(s[0]);
			memcpy(s[i]+l,s[i],l); s[i][l<<1]=0;
			getmin(s[i],l);
		}
		sort(z,z+n,cmp);
		printf("%d
", unique(z,z+n,eq)-z);
	}
}