HDU 1358 Period

E - Period
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status

Practice

HDU 1358
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

题意很简单，就是 给你一个字符串，让你找他的 前缀串 是不是由 某个字符串循环而形成，如果是输出 前缀串的长度 和 循环次数。

我只能说题意很简单，但是这数据就是一脸的懵逼 - -

#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <iostream>
#define   MAX 1000005
using namespace std;
int nexts[MAX];
char s[MAX];
void kmp_nexts(char *s,int *n)
{
    int i=0,len=strlen(s),k=-1;
    n[0]=-1;
    while(i<len)
    {
        if(k==-1||s[i]==s[k])
            n[++i]=++k;
        else
            k=n[k];
    }
}
int main()
{
    int t;
    int times=1;
    while(~scanf("%d",&t)&&t)
    {
        getchar();
        scanf("%s",s);
        kmp_nexts(s,nexts);
        int len=strlen(s);
        printf("Test case #%d\n",times++);
        for(int i=0;i<=t;i++)
        {
            if(nexts[i]==-1||nexts[i]==0)
                continue;

            int k=i-nexts[i];      //这里表示 i 字节前的最小循环节的长度。
                                 // nexts[i] 表示 I这点与之前重复的字符数，那么用i  减去 Nexts[i] 就是 i 这个字节前的循环循环长度了。
            if(i%k==0)       如果 i 除以这个循环长度等于零说明满足题意输出就好。      
               printf("%d %d\n",i,i/k);
        }
        printf("\n");
    }
    return 0;
}