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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#689125#8334. Geneadivse#ML 0ms0kbC++203.2kb2024-10-30 15:27:392024-10-30 15:27:41

Judging History

你现在查看的是最新测评结果

  • [2024-10-30 15:27:41]
  • 评测
  • 测评结果:ML
  • 用时:0ms
  • 内存:0kb
  • [2024-10-30 15:27:39]
  • 提交

answer

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <iomanip>
#define endl '\n'
#define int long long
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define rep2(i,a,b) for(int i=(a);i>=(b);i--)
using namespace std;
template<typename T>
void cc(vector<T> tem) { for (auto x : tem) cout << x << ' '; cout << endl; }
void cc(int a) { cout << a << endl; }
void cc(int a, int b) { cout << a << ' ' << b << endl; }
void cc(int a, int b, int c) { cout << a << ' ' << b << ' ' << c << endl; }
void kuaidu() { ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
inline int max(int a, int b) { if (a < b) return b; return a; }
inline int min(int a, int b) { if (a < b) return a; return b; }
void cmax(int& a, const int b) { if (b > a) a = b; }
void cmin(int& a, const int b) { if (b < a) a = b; }
using PII = pair<int, int>;
using i128 = __int128;

//--------------------------------------------------------------------------------
const int N = 6e4 + 10;
const int M = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 1e16;
int n, m, T;
struct HASH {
    int base = 131;
    int P[2][N], pre[2][N];
    char S[N];
    int Mod[2];
    void init(string s) {//需要保证字符串是从0开始的
        Mod[0] = 998244353;
        Mod[1] = 1e9 + 7;
        int n = s.size();
        for (int i = 1; i <= n; i++) S[i] = s[i - 1];
        P[0][0] = P[1][0] = 1;
        for (int i = 0; i <= 1; i++) {
            for (int j = 1; j <= n; j++) {
                pre[i][j] = (pre[i][j - 1] * base % Mod[i] + S[j]) % Mod[i];
                P[i][j] = P[i][j - 1] * base % Mod[i];
            }
        }
    }
    PII qry(int l, int r) {
        int a, b; l++, r++;
        a = (pre[0][r] - pre[0][l - 1] * P[0][r - l + 1] % Mod[0] + Mod[0]) % Mod[0];
        b = (pre[1][r] - pre[1][l - 1] * P[1][r - l + 1] % Mod[1] + Mod[1]) % Mod[1];
        return { a,b };
    }
}hc[310];
//--------------------------------------------------------------------------------
int k;
HASH tem;
bool check(int x, int l, int r) {
    PII q1 = hc[x].qry(l, r);
    PII q2 = tem.qry(l, r);
    if (q1 != q2) return 1;
    return 0;
}
int dfs(int x) {
    int pr = 0;
    int cnt = 0;
    rep(i, 1, k + 1) {
        int l = pr - 1, r = m;

        while (l + 1 != r) {
            int mid = l + r >> 1;
            if (check(x, pr, mid)) r = mid;
            else l = mid;
        }
        // cout << x << " " << r << endl;
        pr = r + 1;
        if (r >= m) break;
        if (r == m - 1) {
            cnt++;
            break;
        }
        cnt++;
    }
    if (cnt <= k) return 1;
    return 0;
}

signed main() {
    kuaidu();
    T = 1;
    //cin >> T;
    while (T--) {
        int q;
        cin >> n >> q >> m >> k;
        rep(i, 1, n) {
            string s; cin >> s;
            hc[i].init(s);
        }
        rep(i, 1, q) {
            string s; cin >> s;
            tem.init(s);
            int ans = 0;
            rep(j, 1, n) {
                ans += dfs(j);
            }
            cout << ans << endl;
        }

    }
    return 0;
}

Details

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Test #1:

score: 0
Memory Limit Exceeded

input:

6 4 4 1
kaki
kika
manu
nana
tepu
tero
kaka
mana
teri
anan

output:

2
2
1
0

result: