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  • hdu 2609 How many 最小表示法

    How many

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1248    Accepted Submission(s): 486


    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.
     
    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
     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <algorithm>
     4 #include <string.h>
     5 #include <set>
     6 using namespace std;
     7 set<string>ss;
     8 void make(char a[],int x,int l)
     9 {
    10     char b[150];
    11     int i;
    12     for(i=0; i<l; i++)
    13         b[i]=a[x+i>=l?x+i-l:x+i];
    14     b[i]='';
    15     ss.insert(b);
    16 }
    17 void moremin(char a[])
    18 {
    19     int len=strlen(a);
    20     int i,j,k,t;
    21     k=i=0;
    22     j=1;
    23     while(i<len&&j<len&&k<len)
    24     {
    25         int t=a[i+k>=len?i+k-len:i+k]-a[j+k>=len?j+k-len:j+k];
    26         if(!t)k++;
    27         else
    28         {
    29             if(t>0) i+=k+1;
    30             else j+=k+1;
    31             if(i==j)j++;
    32             k=0;
    33         }
    34     }
    35     make(a,(i>j?j:i),len);
    36 }
    37 int main()
    38 {
    39     int n;
    40     char a[150];
    41     while(~scanf("%d",&n))
    42     {
    43         ss.clear();
    44         while(n--)
    45         {
    46             scanf("%s",a);
    47             moremin(a);
    48         }
    49         cout<<ss.size()<<endl;
    50     }
    51 }
    View Code
     
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  • 原文地址:https://www.cnblogs.com/ERKE/p/3832893.html
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