POJ 2406 Power Strings(KMP:找串迴圈節)
Power Strings
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 60362 | Accepted: 24990 |
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
演算法分析:
分析:poj 1961 Period (KMP+最小迴圈節)
程式碼實現:
#include<cstdio> #include<iostream> #include<fstream> #include<algorithm> #include<functional> #include<cstring> #include<string> #include<cstdlib> #include<iomanip> #include<numeric> #include<cctype> #include<cmath> #include<ctime> #include<queue> #include<stack> #include<list> #include<set> #include<map> using namespace std; const int N = 1000002; int nxt[N]; char T[N]; int tlen; void getNext() { int j, k; j = 0; k = -1; nxt[0] = -1; while(j < tlen) if(k == -1 || T[j] == T[k]) { nxt[++j] = ++k; if (T[j] != T[k]) //優化 nxt[j] = k; } else k = nxt[k]; } int main() { int TT; int kase=1; while(scanf("%s",&T)!=-1) { tlen= strlen(T); if(tlen==1&&T[0]=='.') break; getNext(); if(tlen%(tlen-nxt[tlen])==0) printf("%d\n",tlen/(tlen-nxt[tlen])); else printf("1\n"); } return 0; }