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);
}
}