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自然万物都趋向从有序变得无序

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POJ 2406 Power Strings(KMP)

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题目链接:点我

题目

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).

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.

题意:

给你一串字符串,找出它是由某个字符串循环多少次构成的,

思路:

KMP算法,裸题,算出next[len] 即可得出答案,

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 1e6+10;
char s[maxn];
int nextt[maxn];

void getnextt(int len){
    memset(nextt, 0, sizeof(nextt));
    nextt[0] = -1;
    int i = 0;
    int  j = -1;
    while(i < len){
        if(j == -1 || s[i] == s[j]){
            ++i;
            ++j;
            if(s[i] != s[j])
                nextt[i] = j;
            else nextt[i] = nextt[j]; 
        }else  j = nextt[j];
    }
}

int main(){
 //   std::ios::sync_with_stdio(false);
    while(cin>>s && s[0] != '.'){
        int len = strlen(s);
        getnextt(len);
        int ans = len/(len - nextt[len]);
        if(len % (len - nextt[len]) != 0)//特殊情况
            ans = 1;
        printf("%d\n",ans);
    }return 0;
}