P4720

后缀自动机SAM

建 SAM ，对于每个节点维护一个区间 \([l,r]\) 表示这个节点即其儿子能到的最大和最小的原串上的位置

对于每个点，答案为 \(min(r-l+1,len)\) ，因为要保证不能重复

// Author: xiaruize
#ifndef ONLINE_JUDGE
bool start_of_memory_use;
#endif
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
    if (a > b)
        return a;
    return b;
}
int min(int a, int b)
{
    if (a < b)
        return a;
    return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

int n, a, b, c, d;
bool s[N];

struct SAM
{
    struct node
    {
        int len, lnk, nxt[2];
        int mi, mx;
    } s[N];
    int cnt = 1, la = 1;

    void ins(int x, int id)
    {
        int cur = ++cnt, p = la;
        s[cur].len = s[la].len + 1;
        while (p && !s[p].nxt[x])
        {
            s[p].nxt[x] = cur;
            p = s[p].lnk;
        }
        int q = s[p].nxt[x];
        if (!p)
            s[cur].lnk = 1;
        else if (s[p].len + 1 == s[q].len)
            s[cur].lnk = q;
        else
        {
            int r = ++cnt;
            s[r] = s[q];
            s[r].len = s[p].len + 1;
            s[r].mx = 0;
            s[r].mi = n;
            while (p && s[p].nxt[x] == q)
            {
                s[p].nxt[x] = r;
                p = s[p].lnk;
            }
            s[cur].lnk = s[q].lnk = r;
        }
        s[cur].mi = s[cur].mx = id;
        la = cur;
    }
} sam;

vector<int> g[N];

void dfs(int x)
{
    for (auto v : g[x])
    {
        dfs(v);
        sam.s[x].mi = min(sam.s[x].mi, sam.s[v].mi);
        sam.s[x].mx = max(sam.s[x].mx, sam.s[v].mx);
    }
}

void solve()
{
    cin >> a >> b >> c >> d >> n;
    for (int i = 0; i < a; ++i)
    {
        b = (b * c + d) % n;
        s[b + 1] = true;
    }
    rep(i, 1, n) sam.ins(s[i], i);
    rep(i, 2, sam.cnt) g[sam.s[i].lnk].push_back(i);
    dfs(1);
    int res = 0;
    rep(i, 2, sam.cnt) res += max(min(sam.s[i].mx - sam.s[i].mi, sam.s[i].len) - sam.s[sam.s[i].lnk].len, 0ll);
    cout << res << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--)
        solve();
#ifndef ONLINE_JUDGE
    cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
    cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
    return 0;
}