地址：http://acm.hdu.edu.cn/showproblem.php?pid=2609

題目：

How many

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2625????Accepted Submission(s): 1135

Problem Description 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. ?

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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'). ?

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

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Sample Input 4 0110 1100 1001 0011 4 1010 0101 1000 0001 ?

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Sample Output 1 2 ?

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Author yifenfei ?

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Source 奮斗的年代 ?

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Recommend yifenfei ? 思路：最大最小表示法+set 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <set> 5 #include <string> 6 using namespace std; 7 8 #define MP make_pair 9 #define PB push_back 10 typedef long long LL; 11 const double eps=1e-8; 12 const int K=1e6+7; 13 const int mod=1e9+7; 14 15 char sb[205]; 16 //ff為真表示最小，為假表示最大 17 //S串應該為原串復制兩次后的字符串 18 int mx_mi_express(char *S,bool ff,int len) 19 { 20 int i=0,j=1,k; 21 while(i<len&&j<len) 22 { 23 k=0; 24 while(k<len&&S[i+k]==S[j+k]) k++; 25 if(k==len) return i<=j?i:j; 26 if((ff&&S[i+k]>S[j+k]) || (!ff&&S[i+k]<S[j+k])) 27 { 28 if(i+k+1>j) i=i+k+1; 29 else i=j+1; 30 } 31 else if((ff&&S[i+k]<S[j+k]) || (!ff&&S[i+k]>S[j+k])) 32 { 33 if(j+k+1>i) j=j+k+1; 34 else j=i+1; 35 } 36 } 37 return i<=j?i:j; 38 } 39 string tmp; 40 set<string>st; 41 int main(void) 42 { 43 int t,n,len; 44 while(scanf("%d",&n)==1&&n) 45 { 46 st.clear(),tmp.clear(),len=0; 47 for(int i=1,be;i<=n;i++) 48 { 49 scanf("%s",sb); 50 if(!len) 51 { 52 len=strlen(sb); 53 for(int j=0;j<len;j++) 54 tmp+='0'; 55 } 56 for(int j=0;j<len;j++) 57 sb[j+len]=sb[j]; 58 be=mx_mi_express(sb,1,len); 59 for(int j=0;j<len;j++) 60 tmp[j]=sb[j+be]; 61 st.insert(tmp); 62 } 63 printf("%d\n",st.size()); 64 } 65 return 0; 66 }

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轉載于:https://www.cnblogs.com/weeping/p/6670265.html

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