Power Strings 分类： POJ 串...

最新推荐文章于 2024-11-30 21:52:08 发布

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最新推荐文章于 2024-11-30 21:52:08 发布

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原文链接：http://www.cnblogs.com/juechen/p/4721948.html

本文详细介绍了如何解决POJ2406问题，即求解给定字符串s的幂运算中最大的n值，使得s可以表示为a^n的形式。通过构建next数组实现字符串匹配，解决此复杂度较高的问题。

Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

POJ 2406
Appoint description:
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.
开始的时候数组开了，orz

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define RR freopen("output.txt","r",stdoin)
#define WW freopen("input.txt","w",stdout)
typedef long long LL;

using namespace std;

const int MAX = 1000010;

char s[MAX];

int next[MAX];

int len;

void Build_next()
{
    int i=0,j=-1;
    len=strlen(s);
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        {
            j=next[j];
        }
    }
}

int main()
{
    while(scanf("%s",s)&&strcmp(s,"."))
    {
        Build_next();
        int ans=len-next[len];
        if(len%ans)
        {
            printf("1\n");
        }
        else
        {
            printf("%d\n",len/ans);
        }
    }
    return 0;
}