Period hdu-1385 ——mystical_curve

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. 

InputThe 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. 
OutputFor 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：对于一个指定长度的字符串，求其前缀能够表示为长度为A的字符串连续表示k>1次,按照前缀长度递增的顺序输出前缀字符的长度i和k

解题思路：kmp的运用

// 
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
#define pb push_back
#define mk make_pair
typedef long long ll;
using namespace std;
const int inf = 1e6+100;
char s[inf];
int n,t=0,nex[inf];
void fun1()//构造next数组
{
    nex[0]=-1;
    int k=-1,j=0;
    while(j<n)
    {
        if(k==-1||s[j]==s[k]){
            j++;k++;
            nex[j]=k;
        }else
        k=nex[k];
    }
}
void fun2()//遍历一遍next数组，病按照结果的要求输出
{
    int k;
    printf("Test case #%d\n",++t);
    for(int i=1;i<n;i++)
    {
        k=i-nex[i];
        //cout<<i<<" "<<k<<endl;
        if(s[i]==s[nex[i]]&&(i+1)%k==0){
            cout<<i+1<<" "<<(i+1)/k<<endl;
        }
    }
    cout<<endl;
}
int main()
{
    while(cin>>n,n!=0)
    {
        scanf("%s",s);
        fun1();//构造next数组
        fun2();//遍历一遍next数组，病按照结果的要求输出
    }
}