Period （poj 1961&&hdu 1358）KMP

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Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 13504		Accepted: 6365

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 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 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

Source

Southeastern Europe 2004

题意：求长度为i（2<=i<=N）的前缀，若前缀是一个周期串，则输出长度i和它的最大周期；要找出所有满足条件的。

思路：next[i]数组里面存的是i位置前 字符串的相同前缀和后缀的最大长度，若它是一个周期串，那它必满足i%（i-next[i]）==0.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 1000010
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int next[MAXN];
char str[MAXN];
int N;

void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    while (i<N)
    {
        if (j==-1||str[i]==str[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}

void KMP()
{
    for (int i=1;i<=N;i++) //枚举所有长度，看它满不满足要求
    {
        if (!(i%(i-next[i]))&&(i/(i-next[i]))>1)//满足就输出
            printf("%d %d\n",i,i/(i-next[i]));
    }
}

int main()
{
    int cas=1;
    while (scanf("%d",&N)&&N)
    {
        scanf("%s",str);
        get_next();
        printf("Test case #%d\n",cas++);
        KMP();
        printf("\n");
    }
    return 0;
}
/*
3
aaa
12
aabaabaabaab
0
*/