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P - How many
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.
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.
InputThe 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‘).
OutputFor 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
#include<iostream> #include<set> #include<map> #include<vector> #include<string> #include<algorithm> #include<cstring> using namespace std; #define MAXN 10002 /* 题目相当于求所有字符串中 能通过相互循环位移得到的字符串数目 对每个字符串求最小表示法,然后加入到set中 */ string str; set<string> S; int GetMin(string s,int len) { int i=0,j=1,k=0; while(i<len&&j<len&&k<len) { if(s[(i+k)%len]==s[(j+k)%len]) k++; else if(s[(i+k)%len]>s[(j+k)%len]) { i = i+k+1; k = 0; } else { j = j+k+1; k = 0; } if(i==j) j++; } return min(i,j); } int main() { int n; string tmp; tmp.reserve(101); while(scanf("%d",&n)!=EOF) { for(int i=0;i<n;i++) { tmp.clear(); cin>>str; int pos = GetMin(str,str.size()),L = str.size(); for(int j=pos,cnt=0;cnt<L;j=(j+1)%L,cnt++) { tmp.push_back(str[j]); } S.insert(tmp); } cout<<S.size()<<endl; S.clear(); } }
P - How many
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