SDUTOJ2475 Period & 2476 Power Strings

版权声明：iQXQZX https://blog.csdn.net/Cherishlife_/article/details/84980933

通过前缀表 求母串的子串的周期循环次数

例题：SDUTOJ2476Period

https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/2710/pid/2476

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define N 1000001
char s[N];
int nextt[N];
void get_next(char *s)
{
    int i = 0, j = -1;
    int len = strlen(s);
    nextt[0] = -1;
    while (i <= len)
    {
        if (j == -1 || s[i] == s[j])
        {
            i++;
            j++;
            nextt[i] = j;
        }
        else
            j = nextt[j];
    }
}
int main()
{
    int n;
    int cnt = 0;
    while (~scanf("%d", &n))
    {
        if (n == 0)
            break;
        scanf("%s", s);
        get_next(s);
        cnt++;
        printf("Test case #%d\n", cnt);
        for (int i = 2; i <= n; i++) // 串的 包含前i个元素的子串的周期循环次数
        {
            if (i % (i - nextt[i]) == 0 && nextt[i]) // i - nextt[i]为一个周期长度
                printf("%d %d\n", i, i / (i - nextt[i]));
        }
        printf("\n");
    }
    return 0;
}

例题：SDUTOJ2475 Power Strings

https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/2710/pid/2475

​
#include <bits/stdc++.h>
using namespace std;
#define N 1000001
char s[N];
int nextt[N];
void get_next(char *s)
{
    int i = 0, j = -1;
    int len = strlen(s);
    nextt[0] = -1;
    while (i <= len)
    {
        if (j == -1 || s[i] == s[j])
        {
            i++;
            j++;
            nextt[i] = j;
        }
        else
            j = nextt[j];
    }
}
int main()
{
    int n;
    int cnt = 0;
    while (~scanf("%s", s))
    {
        if (s[0] == '.')
            break;
        n = strlen(s);
        get_next(s);
        if (n % (n - nextt[n]) == 0 && nextt[n]) // 能通过前缀表找到周期 则输出
            printf("%d\n", n / (n - nextt[n])); // 输出最大周期循环次数  n - nextt[n]为一个周期长度
        else
            printf("%d\n", 1); // 如果通过前缀表找不到周期则输出 1 
    }
    return 0;
}

​