Period（kmp）

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4437    Accepted Submission(s): 2145

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

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

 题解：看了半天题意，，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，
   这个k需要大于1。于是我们可以想到，next【i】的值就是长度，当i%next[i]==0是开始匹配i/next[i]便是重复度k;

 

用了strlen就超时了。。。看来题目中给的N很有必要啊；

代码贴上：

#include<stdio.h>
const int MAXN=1000010;
int N;
char m[MAXN];
int next[MAXN];
void getnext(){
    int i=0,j=-1;
    next[i]=j;
    while(i<N){
        if(j==-1||m[i]==m[j]){
            i++;j++;
            next[i]=j;
        }
        else j=next[j];
    }
}
void print(int x){
     for(int i=1;i<=x;i++){
         if(next[i]!=0&&i%(i-next[i])==0)printf("%d %d\n",i,i/(i-next[i]));
     }
}
int main(){
    int flot=0;
    while(~scanf("%d",&N),N){
        scanf("%s",m);
        getnext();
        //for(int i=0;i<=N;i++)printf("%d ",next[i]);puts("");
        //if(flot)puts("");
        printf("Test case #%d\n",++flot);
        print(N);
        puts("");
    }
    return 0;
}

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace  std;
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
int p[MAXN];
char s[MAXN];
int N;
void getp(){
    int i=0,j=-1;
    p[0]=-1;
    while(i<N){
        if(j==-1||s[i]==s[j]){
            i++;j++;p[i]=j;
            if(s[i]!=-1&&i%(i-p[i])==0&&i/(i-p[i])>1){
                printf("%d %d\n",i,i/(i-p[i]));
            }
        }
        else j=p[j];
    }
}
/*
void kmp(int& ans){
    getp();
    int j=0,i=0;
    while(i<N){
        if(j==-1||s[j]==m[i]){
            i++;j++;
            if(j==M){
                ans=i-j+1;
                return ;
            }
        }
        else j=p[j];
    }
}
*/
int main(){
    int kase=0;
    while(~scanf("%d",&N),N){
        scanf("%s",s);
        printf("Test case #%d\n",++kase);
        getp();
        puts("");
    }
    return 0;
}