uva 719 glass beads

学习了周源的最小表示法，很快解决了，时间复杂度是O(n)

 

#include <cstdio>#include <cassert>#include <cstring>#include <string>#include <algorithm>using namespace std;int MinExp(const string &str) { string s = str + str; int i = 0, j = 1, k = 0; while(i < str.size() && j < str.size() && k < str.size()) { if(s[i + k] == s[j + k]) { ++k; } else if(s[i + k] < s[j + k]) { j = max(i + 1, j + k + 1); k = 0; } else { i = max(j + 1, i + k + 1); k = 0; swap(i, j); }#ifdef _DEBUG printf("i = %d, j = %d, k = %d/n", i, j, k);#endif } return i;}int main() { int cs; scanf("%d", &cs); while(cs--) { char buf[10240]; scanf("%s", buf); printf("%d/n", MinExp(buf) + 1); } return 0;}