HDU1358（KMP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1358

主要考查KMP算法next数组本质的理解。len - len[len]为循环节的大小~~~

代码如下：

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1000002;
char str[N];
int next[N];

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

int main()
{
    int n;
    int k = 0;
    while(scanf("%d", &n) != EOF && n != 0)
    {
        scanf("%s", str);
        get_next(n);
        k++;
        printf("Test case #%d\n", k);
        for(int i = 2; i <= n; i++)
        {
            int x = i - next[i];
            if((i % x == 0) && i / x != 1)
            {
                printf("%d %d\n", i, i / x);
            }
        }
        printf("\n");
    }
    return 0;
}

参考别人的代码：

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1000002;
char str[N];
int  next[N];

void get_next(int len)
{
    int i;
    int j = 0;
    for(next[1] = 0, i = 2; i <= len; i++)
    {
        while(j && str[i] != str[j + 1])
        {
            j = next[j];
        }
        if(str[i] == str[j + 1])
        {
            j++;
        }
        next[i] = j;
    }
}

int main()
{
    int n;
    int k = 1;
    while(scanf("%d", &n) != EOF && n != 0)
    {
        scanf("%s", str + 1);
        get_next(n);
        printf("Test case #%d\n", k++);
        for(int i = 2; i <= n; i++)
        {
            if(i % (i - next[i]) == 0 && i / (i - next[i]) != 1)
            {
                printf("%d %d\n", i, i / (i - next[i]));
            }
        }
        printf("\n");
    }
    return 0;
}