poj2406——Power Strings(KMP)

最新推荐文章于 2018-08-09 11:21:37 发布

原创最新推荐文章于 2018-08-09 11:21:37 发布 · 570 阅读

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本文介绍了一个经典的字符串处理问题，即如何判断一个字符串是否可以由另一个较短字符串重复组成，并求出该较短字符串的最大重复次数。通过使用KMP算法进行高效解决。

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Power Strings

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 25855Accepted: 10829

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output

For each s you should print the largest n such that s = a^n for some string a.
Sample Input

abcd
aaaa
ababab
.
Sample Output
1
4
3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source

Waterloo local 2002.07.01

解析：

kmp的经典应用。。。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;

char s[1000010];
int next[1000010];
int len;

void read()
{
    freopen("poj2406.in","r",stdin);
    freopen("poj2406.out","w",stdout);
}

int get_next()
{
    int i=0,j=-1;
    next[i]=j;
    while(i<len)
    {
        if(j==-1 || s[i]==s[j])
        {
            i++;j++;
            next[i]=j;
        } else
        j=next[j];
    }
    i=len-j;
    if(len%i==0)return len/i;
    else return 1;
}

int main()
{
    read();
    while(scanf("%s\n",s)!=EOF)
    {
        if(s[0]=='.')break;
        len=strlen(s);
        printf("%d\n",get_next());
    }
    return 0;
}