POJ 2406 Power Strings（next数组应用）

G - Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题意:给你一个字符串，让你求这个字符串是由多少个相同的子串构成的，要求子串数量尽量多；

思路：如果你随便找一个循环构成的子串，对他手动构造kmp的next数组，你会发现，出了第一个周期，其余的周期next数组都是增加的，而且这个周期肯定是最短周期

假如由k个周期，next[n]肯定就是k-1个周期的字母个数，用n - next[n]就求出第一个周期的长度，如果能用总长度n对n - next[n]正好取膜，说明他是第一个周期。。

如果不能整除，可能是这个循环子串前面加了一些东西，或者就只有一个。。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 1e6 + 5;int next[maxn];char str[maxn];int compute(int *next, char *p){    int l = strlen(p);    int k = 0;    next[0] = next[1] = 0;    for(int i = 2; i <= l; i++)    {        while(k && p[k] != p[i-1])            k = next[k];        if(p[k] == p[i-1])            k++;        next[i] = k;    }    return l-next[l];}int main(){    while(~scanf("%s",str) && strcmp(str,"."))    {        int n = strlen(str);        int k = compute(next,str);        if(n%k == 0)            cout << n/k << endl;        else            cout << 1 << endl;    }    return 0;}