Power Strings 分类： POJ 串 2015-07-31 19:05 8人阅读 评论(0) 收藏

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
");
        }
        else
        {
            printf("%d
",len/ans);
        }
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。