CC

自然万物都趋向从有序变得无序

0%

POJ 1961 Period(KMP)

发表于 更新于 分类于 Valine：

题目连接:点我

 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 A K , 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

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

题意:

给定一个字符串，问这个字符串的所有前缀中，有哪些前缀可以由某个串重复k次组成， 并且这个k最大是多少。这个k需要大于1。

思路:

KMP算法,对于每次匹配成功,我们都判断是否

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn = 1e6+10;
char s[maxn];
int nextt[maxn];

void getnextt(int len){
        std::ios::sync_with_stdio(false);

    int i = 0, j = -1;
    nextt[0] = -1;
    while(i<len){
        if(j == -1 || s[i] == s[j]){
            i++; j++;
            nextt[i] = j;
            if(i % (i - j) == 0 && i/(i-j )> 1)// i - j 为循环节长度,
                cout<<i<<' '<<i/(i-j)<<endl;
        }else  j = nextt[j];
    }
}

int main(){
    int n;
    std::ios::sync_with_stdio(false);//解除输入限制,可以提高 cin 的速度
    int kase = 0;
    while(cin>>n && n){
        cin>>s;
        memset(nextt, 0, sizeof(nextt));
        cout<<"Test case #"<<++kase<<endl;
        getnextt(n);
        cout<<endl;
    }return 0;
}