FZU-1901-Period 2(KMP)

链接:

https://vjudge.net/problem/FZU-1901

题意:

For each prefix with length P of a given string S,if
S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

思路:

求可能的循环节长度.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e6+10;
const int MOD = 1e4+7;

char a[MAXN];
int Next[MAXN], Res[MAXN];
int n;

void GetNext(char *s)
{
    int len = strlen(s);
    int j = 0, k = -1;
    Next[0] = -1;
    while (j < len)
    {
        if (k == -1 || s[j] == s[k])
        {
            ++j;
            ++k;
            Next[j] = k;
        }
        else
            k = Next[k];
    }
}

int main()
{
    int t, cnt = 0;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%s", a);
        int len = strlen(a);
        GetNext(a);
        int ans = 0;
        int p = len;
        while (p > 0)
        {
            Res[++ans] = len-Next[p];
            p = Next[p];
        }
        printf("Case #%d: %d
", ++cnt, ans);
        printf("%d", Res[1]);
        for (int i = 2;i <= ans;i++)
            printf(" %d", Res[i]);
        puts("");
    }

    return 0;
}