给一个串,问有多少和它长度相同的串,使得LCS为l - 1。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
const int N = 100005;
int n, m, p, q, s1[N], s2[N]; char s[N]; LL ans;
int main()
{
scanf ("%d%d", &n, &m);
scanf ("%s", s + 1);
ans = n * (m - 1);
Rep(i, 1, n - 1) {
s1[i] = s1[i - 1] + (s[i] != s[i + 1]);
ans += s1[i] * (m - 1);
}
Dwn(i, n, 2) {
s2[i] = s2[i + 1] + (s[i] != s[i - 1]);
ans += s2[i] * (m - 1);
}
Rep(i, 1, n - 1) {
p = max(0, p - 2);
while (i + p + 2 <= n && s[i + p] == s[i + p + 2]) p ++;
if (s[i] != s[i + 1]) ans -= p + 1;
swap(p, q);
}
printf("%I64d\n", ans);
return 0;
}