2019年ccpc秦皇岛赛区（J题）重现赛1010

题意：

求的是后缀的循环节长度B和循环节出现的总长度(包含不完整循环节)A，给定a和b求，最大的aA-bB，显然kmp反过来后跑一下求出next数组，枚举前缀，那么i就是循环节出现的总长度，而i−next[i]就是循环节长度了，直接更新最大值，a，b和ans都需要ll，答案可能有负数，初值要设为-1e18。

代码：

// nuoyanli
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define inf 0x3f3f3f3f
#define lson l, mid, root << 1
#define rson mid + 1, r, root << 1 | 1
#define min(a, b) a > b ? b : a
#define max(a, b) a < b ? b : a
#define ll long long int
#define mod 998244353
using namespace std;

const int N = 1e7 + 10;
char s[N], s1[N];
ll len, t, nxt[N];
inline void get_next(int len) {
  int i, j;
  nxt[0] = -1;
  for (i = 0; i < len; ++i) {
    j = nxt[i];
    while (j != -1 && s[j] != s[i])
      j = nxt[j];
    nxt[i + 1] = j + 1;
  }
}
signed main() {
  ll a, b;
  while (~scanf("%lld %lld", &a, &b)) {
    scanf("%s", s1);
    int len2 = strlen(s1) - 1;
    int tot = 0;
    for (int i = len2; i >= 0; i--) {
      if (s1[i] == '.')
        break;
      s[tot++] = s1[i];
    }
    len = strlen(s);
    get_next(tot);
    ll ans = -1e18;
    for (int i = 1; i <= tot; i++) {
      ans = max(ans, a * i - b * (i - nxt[i]));
    }
    printf("%lld\n", ans);
  }
  return 0;
}