HDU 1358 Period

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3749    Accepted Submission(s): 1841

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 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

3aaa12aabaabaabaab0

 

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

 

呃。。

题意大意就是，从第二个位置开始，如果前面有相同的循环串，则输出位置此时的位置下表，并且输出循环串的个数。

思路：跑个KMP，（只有某个位置的移动距离与当前位置的相差的值）为当前位置的1/m之一时，并且next数组里面的值不为0时，才有循环串出现。（窝时这样理解的，不知道对不对）。。

上代码。。

#include <stdio.h>#include <cstring>#include <cmath>#include <cstdlib>#include <iostream>#include <algorithm>#include <queue>#include <map>#include <vector>using namespace std;#define mod 1000000007#define inf 0x7f7f7f7fint Next[1000005];char s[1000005]; //看好题目，，数组开100W，我开了10WRE了好多次。。int l;void getNext(){    int i,j;    i=0;    j=-1;    Next[i]=j;    while(i<l)    {        if(j==-1 || s[i]==s[j])        {            i++;            j++;            Next[i]=j;        }        else            j=Next[j];    }}int main(){    int n;    int Case=0;    while(cin>>n,n)    {        scanf("%s",s);        l=strlen(s);        memset(Next,0,sizeof(Next));        getNext();        printf("Test case #%d\n",++Case);        int k;        // for(k=0;k<l;k++)        //     cout<<Next[k]<<endl;        //     cout<<Next[l]<<endl;        for(k=1; k<l; k++)        {            if(Next[k+1]!=0 && (k+1)%(k+1-Next[k+1])==0)  //只有next数组不为0时才有循环节出现，，循环节的长度就是当前的长度减去next数组每次移动的长度。                cout<<k+1<<" "<<(k+1)/(k+1-Next[k+1])<<endl;        }        cout<<endl;    }    return 0;}

所以，求出