Ultraviolet A 1358-generator (KMP + expected)

Link to the question: Ultraviolet A 1358-generator

Given N, it indicates there are n characters, and then given a string S, the string is empty at the beginning, and now a 1 ~ N is added to the end of the string, and the string contains s as the expected number of times the substring is generated.

Solution: first, we need to pre-process s to find out the mismatch array.

The definition of DP [I] indicates that the end part matches the expected number of times required by the I s string. Each enumeration may contain 1 ~ N. For the S string, I + 1 must be a definite character, so other characters certainly do not match.
Assuming that the K character is generated and the K character is not equal to s [I + 1], we can determine the number of matching Characters Based on the S mismatch array, (similar to the KMP matching problem). Suppose there are j matching characters, that is to say, we need to re-generate DP [I]-DP [J] times from matching J to matching I (expected ).
So f (I) (from matching I-1 to matching I need to generate the expected number of times) There is a formula f (I) = 1 + Σ I = 1N (DP [I? 1]? DP [lose (k)]) + n? 1nf (I) (lose (k) is the number of matched characters when the corresponding generated character is K)
DP [I] = DP [I-1] + f (I)

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 20;int len, jump[maxn];void get_jump(char* s) {    int p = 0;    len = strlen(s+1);    for (int i = 2; i <= len; i++) {        while (p && s[p+1] != s[i])            p = jump[p];        if (s[p+1] == s[i])            p++;        jump[i] = p;    }}ll solve () {    int n;    ll dp[maxn];    char s[maxn];    scanf("%d%s", &n, s+1);    get_jump(s);    dp[0] = 0;    for (int i = 1; i <= len; i++) {        ll& ans = dp[i];        ans = dp[i-1] + n;        for (int j = 0; j < n; j++) {            if (s[i] == ‘A‘ + j)                continue;            int p = i-1;            while (p && s[p+1] != j + ‘A‘)                p = jump[p];            if (s[p+1] == j + ‘A‘)                p++;            ans += dp[i-1] - dp[p];        }    }    return dp[len];}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        printf("Case %d:\n%lld\n", kcas, solve());        if (kcas < cas)            printf("\n");    }    return 0;}

Ultraviolet A 1358-generator (KMP + expected)