题意：给你一个字符串向量，问你有中间多少个串循环相等（循环相等的意思就是 “ab”，“ba”，经过移位以后还相等的字符串）
解题思路：KMP，这种字符串有一个性质，就是 如果 a的长度等于b的长度，且 能在 a+a 中匹配到b，则可以断定这两个字符串循环相等
解题代码：
#include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <string> #include <iostream> #include <sstream> #include <map> #include <set> #include <queue> #include <stack> #include <fstream> #include <numeric> #include <iomanip> #include <bitset> #include <list> #include <stdexcept> #include <functional> #include <utility> #include <ctime> using namespace std; #define PB push_back #define MP make_pair #define REP(i,n) for(i=0;i<(n);++i) #define FOR(i,l,h) for(i=(l);i<=(h);++i) #define FORD(i,h,l) for(i=(h);i>=(l);--i) typedef vector<int> VI; typedef vector<string> VS; typedef vector<double> VD; typedef long long LL; typedef pair<int,int> PII; int kmp(string a, string b) { int pi[100]; int m = b.size(); memset(pi,0,sizeof(pi)); pi[0] = -1; int k = -1 ; for(int i = 1 ;i < m ; i ++) { while(k >= 0 && b[k+1] != b[i]) k = pi[k]; if(b[k+1] == b[i]) k = k +1; pi[i] = k; } int n = a.size(); int q = -1 ; for(int i = 0;i < n;i ++) { while(q >= 0 && b[q+1] != a[i]) q = pi[q]; if(b[q+1] == a[i]) q = q +1; if(q == m -1) { return 1; } } return 0 ; } class FoxAndWord { public: int howManyPairs(vector <string> words) { int k = words.size(); int sum = 0 ; for(int i = 0;i < k ;i ++) for(int j = i + 1;j < k; j ++) { string temp = words[i] + words[i]; if(kmp(temp,words[j])&& words[i].size() == words[j].size()) { sum ++ ; } } return sum ; } };