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POJ1961[KMP 失配函数]
Period
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 16776 | Accepted: 8077 |
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 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.
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
Source
Southeastern Europe 2004
求每个前缀的最短循环节
错位的部分是个循环节
即i%(i-f[i])==0
再次想想从0开始问题
跳到的那个位置之前的所有字符与当前失配字母前的相同数量个字母是相匹配的
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=1e6+5;int n,f[N];char p[N];int main(){ int cas=0; while(scanf("%d",&n)!=EOF&&n){ scanf("%s",p); f[0]=f[1]=0; for(int i=1;i<n;i++){ int j=f[i]; while(j&&p[j]!=p[i]) j=f[j]; f[i+1]=p[j]==p[i]?j+1:0; } printf("Test case #%d\n",++cas); for(int i=2;i<=n;i++) if(f[i]>0&&i%(i-f[i])==0) printf("%d %d\n",i,i/(i-f[i])); printf("\n"); }}
POJ1961[KMP 失配函数]
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