[NOI 2014] Zoo: KMP algorithm, recursive

Again the wrong question ... This problem does not change. Fortunately, it's not too far. I decided to add the "test instructions brief" to the later writing.

Test instructions: Give a string s, length L not more than 1 million, remember all is both the substring s[0..i] prefix, but also suffix, and the prefix, the number of non-overlapping strings is num[i], num[0], num[1], ..., num[l-1] The result of the modulus of a given large prime number. Multiple sets of data.

The background is given in the surface: KMP algorithm. So I took "quantity" as the "maximum length".

In the "Noi Guide" training course to listen to the teacher said, but did not understand ... and took it out and thought about it.

If the "prefix, suffix does not overlap" is not considered and is not equal to the entire substring, how to find the number of such strings. When you run the KMP, you push an array of CNT, which indicates how many successful matches you've got from I to the next array, that's what you asked for. Because each jump is equivalent to finding a prefix = suffix; cnt[0] = 0.

If we add this restriction.

Run the KMP first, and find the next and the CNT. Run again, just add j*2 >= i in the condition of the jump, and update the answer after each stop. Note that CNT does not count S[0..J], so this is +1.

Aside from this problem, Next[i] is defined as a true prefix equal to the maximum length of the suffix a little awkward ... After jumping out of the while loop, j=0 can either represent a successful match to s[0] or it may mean that there is no match. Define Next[i] as the position of the next comparison after mismatch at I, and place next[0] =-1 may be better.

#include <cstdio> #include <cstring> #include <cassert> using namespace std;
typedef long Long LL;
const int max_l = 1e6, M = 1e9+7;

Char s[max_l+1];

    int solve () {static int next[max_l], cnt[max_l];//cnt[i] ~ substring S[0..I] How many true prefixes are equal to the suffix int L = strlen (s);
        for (int i = 1, j = 0; i < L; ++i) {while (J && s[i]! = s[j]) j = next[j-1];
            if (s[i] = = S[j]) {cnt[i] = Cnt[j] + 1;
        ++j;
        } else Cnt[i] = 0;
    Next[i] = j;

    } ll ans = 1; for (int i = 1, j = 0; i < L; ++i) {while (J && (s[i]! = S[j] | | j*2 >= i)) j = next[j-
        1];
            if (s[i] = = S[j] && j*2 < i) {ans = (ans * (cnt[j]+2))% M;
        ++j;
}} return ans;
    } int main () {int n;
    scanf ("%d", &n);
        while (n--) {scanf ("%s", s);
    printf ("%d\n", Solve ());
} return 0; }