PeriodTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3443Accepted Submission(s): 1727
Problem Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK, that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.
Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3aaa12aabaabaabaab0
Sample Output
Test case #12 23 3Test case #22 26 29 312 4
题意： 从字符串第二个位置开始，前面的字符是否是循环的字符串，如果是，，输出当前位置及其循环的个数。
又是kmp的getnext（）函数循环节的考察
#include<stdio.h>#include<iostream>#include<math.h>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<vector>#include<string.h>#include<queue>#include<stack>#include<set>#include<map>using namespace std;char b[1000500];int Next[1000500];void get_next(char b[], int m){int i = 0,j = -1;memset(Next,0,sizeof(Next));Next[0] = -1;while (b[i]){if (j == -1 || b[i] == b[j]){++i;++j;Next[i] = j;}elsej = Next[j];}}int main(){int cases = 1, n, m, i, j;while (scanf("%d",&n)!=EOF && n){scanf("%s",b);get_next(b, n);printf("Test case #%d\n",cases++);for (int i = 2; b[i-1]; i++){int t = Next[i];int s = i – t;if (i % s == 0 && i/s>1){printf("%d %d\n",i,i/s);}}printf("\n");}return 0;}
有一种缘，放手后成为风景，有一颗心，坚持中方现真诚。
