Power strings (poj 2406 KMP)

Language:DefaultPower strings Time limit:3000 Ms   Memory limit:65536 K Total submissions:33205   Accepted:13804 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 shoshould print the largest N such that S = a ^ N for some string. Sample Input abcdaaaaababab. Sample output 143 Hint This problem has huge input, use scanf instead of CIN to avoid time limit exceed. Source Waterloo local 2002.07.01

The length of a string s cannot exceed 10 ^ 6. Calculate the maximum n so that S is connected by N identical strings a, for example: "ababab" is composed of N = 3 "AB" connections. "aaaa" is composed of N = 4 "A" connections, "ABCD" is connected by n = 1 "ABCD.

Theorem: if the length of S is Len, S has a circular substring. if and only, Len can be divisible by Len-next [Len, the shortest loop substring is s [Len-next [Len]

Idea: Use the KMP algorithm to obtain the next feature vector of a string. If Len can be divisible by Len-next [Len, the maximum number of cycles N is Len/(LEN-next [Len]); otherwise, it is 1.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int N;int nextval[1000010];char str[1000010];int get_nextval(){    int i=0;    int len=strlen(str);    int j=-1;    nextval[i]=-1;    while (i<len)    {        if (j==-1||str[i]==str[j])        {            i++;            j++;            nextval[i]=j;        }        else            j=nextval[j];    }    if ((len)%(len-nextval[len])==0)        return len/(len-nextval[len]);    else        return 1;}int main(){    while (scanf("%s",str))    {        if (str[0]=='.')            return 0;        printf("%d\n",get_nextval());    }    return 0;}

Power strings (poj 2406 KMP)