poj 2406 Power Strings(KMP)

题目链接：http://poj.org/problem?id=2406

思路分析：

1.理解Kmp算法的next数组的意义；

2.对于字符A[i]，i-next[i]等价于在字符串中存在一个长度为i-next[i]的重复子串；

3.当 i % (i - next[i]) == 0 等价于字符串由 (i/(i-next[i])) 个长度为 i - next[i]的子串连接组成;

 

代码： 

#include <iostream>
#include <string>
using namespace std;

const int MAX_N = 1000000 + 10;
char S[MAX_N];
int Next[MAX_N];

int GetNext(char P[], int next[])
{
    int i = 0, j = -1;
    int len = strlen(P);

    if (len <= 0)
        return 0;

    next[0] = -1;
    while (i <= len)
    {
        if (j == -1 || P[i] == P[j])
            next[++i] = ++j;
        else
            j = next[j];
    }
}

int main()
{
    int len;
    int lenSubstr;

    while (scanf("%s", S) != EOF && strcmp(S, "."))
    {
        len = strlen(S);
        GetNext(S, Next);
        lenSubstr = len - Next[len];

        if (len % lenSubstr == 0)
            printf("%d\n", len / lenSubstr);
        else
            printf("1\n");
    }

    return 0;
}