KMP循环节 在icpc

2019 China Collegiate Programming Contest Qinhuangdao Onsite J. MUV LUV EXTRA

由题易得，要求这个数的小数部分的\(S=a×循环长度−b×循环节的长度\),让这个S尽可能的大。
又因为对于循环长度我们可以用kmp算法来求出最小循环节，所以我们可以枚举循环长度去找最小的S.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define pii pair<int, int>
#define all(a) a.begin(), a.end()
#define debug(x) cout << #x << " = " << x << endl
#define sz(a) ((int)a.size())
const int mod = 998244353, N = 2e7 + 10, G = 3;
ll ne[N];
void get_Next(string s, ll next[])
{
    int j = 0;
    next[0] = 0; 
    for (int i = 1; i < s.length(); i++)
    { 
        while (j > 0 && s[i] != s[j])
            j = next[j - 1]; 
        if (s[i] == s[j])
            j++;     
        next[i] = j; 
    }
}

void solve()
{
    ll a, b;
    cin >> a >> b;
    string s;
    cin >> s;
    ll l = s.length();
    reverse(all(s));
    get_Next(s, ne);
    ll ans = -1e18;
    for (int i = 0; i < l; i++)
    {
        if (s[i] == '.')
            break;
        ans = max(ans, 1ll * a * (i + 1) - 1ll * b * (i + 1 - ne[i]));
    }
    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t = 1;
    while (t--)
    {
        solve();
    }
    return 0;
}