zoukankan      html  css  js  c++  java
  • HDU2609 How many —— 最小表示法

    题目链接:https://vjudge.net/problem/HDU-2609

    How many

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3272    Accepted Submission(s): 1457


    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
     
    Author
    yifenfei
     
    Source
     
    Recommend
    yifenfei
     
     
     
    题解:
    1.求出每个字符串的最小表示法。
    2.对所有字符串的最小表示进行排序,然后统计。
     
     
    代码如下:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <string>
     6 #include <vector>
     7 #include <map>
     8 #include <set>
     9 #include <queue>
    10 #include <sstream>
    11 #include <algorithm>
    12 using namespace std;
    13 typedef long long LL;
    14 const double eps = 1e-6;
    15 const int INF = 2e9;
    16 const LL LNF = 9e18;
    17 const int MAXN = 1e4+10;
    18 
    19 char s[MAXN][110], tmp[110];
    20 
    21 int cmp(const void *a,const void *b)
    22 {
    23      return (strcmp((char*)a,(char*)b));
    24 }
    25 
    26 int getmin(char *s, int len) //返回最小表示法的始端
    27 {
    28     int i = 0, j = 1, k = 0;
    29     while(i<len && j<len && k<len)
    30     {
    31         int t = s[(i+k)%len]-s[(j+k)%len];
    32         if (!t) k++;
    33         else
    34         {
    35             if (t>0) i += k+1;
    36             else j += k+1;
    37             if (i==j) j++;
    38             k = 0;
    39         }
    40     }
    41     return i<j?i:j;
    42 }
    43 
    44 int main()
    45 {
    46     int n;
    47     while(scanf("%d", &n)!=EOF)
    48     {
    49         for(int i = 1; i<=n; i++)
    50         {
    51             scanf("%s", tmp);
    52             int len = strlen(tmp);
    53             int k = getmin(tmp, len);
    54             for(int j = 0; j<len; j++)
    55                 s[i][j] = tmp[(k+j)%len];
    56             s[i][len] = 0; //!!
    57         }
    58         qsort(s+1, n, sizeof(s[1]), cmp);
    59 
    60         int ans = 0;
    61         for(int i = 1; i<=n; i++)
    62             if(i==1 || strcmp(s[i], s[i-1]))
    63                 ans++;
    64 
    65         printf("%d
    ", ans);
    66     }
    67 }
    View Code
  • 相关阅读:
    Newtonsoft.Json 把对象转换成json字符串
    分页总页数计算方法 所有分页通用
    好用的Markdown编辑器一览 readme.md 编辑查看
    jquery bootgrid 一个很好的 数据控件,可用于任何语言
    史上最详细“截图”搭建Hexo博客——For Windows
    史上最详细“截图”搭建Hexo博客并部署到Github
    正则表达式30分钟入门教程
    [前端插件]为自己的博客增加打赏功能
    css浮动
    MetaWeblog API调用
  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/7894889.html
Copyright © 2011-2022 走看看