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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#219541#5666. Repetitive Elementsshen0628#AC ✓5ms3512kbC++173.0kb2023-10-19 16:07:452023-10-19 16:07:45

Judging History

你现在查看的是最新测评结果

  • [2023-10-19 16:07:45]
  • 评测
  • 测评结果:AC
  • 用时:5ms
  • 内存:3512kb
  • [2023-10-19 16:07:45]
  • 提交

answer

// #pragma GCC optimize("03,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>
#include <queue>
#include <deque>
#include <bitset>
#include <set>
#include <unordered_set>
#include <utility>
#include <numeric>
#include <iomanip>
#include <stack>
#include <list>
#include <bitset>
#include <sstream>
#include <random>
#include <stdlib.h>
#include <time.h>
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
#define MXN 2000005
#define cl(x) (x << 1)
#define cr(x) ((x << 1) | 1)
#define SZ(x) (int)x.size()
#define PB push_back
#define lowbit(x) (x & (-x))
#define NO_TAG false
#define P1 75577
#define P2 12721
#define MOD1 998244353
#define MOD2 1000000007

using namespace std;

// mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());

// int rand_int(int lb, int ub) {
//     return uniform_int_distribution<int>(lb, ub)(gen);
// }

int n;
string s;
vector<pair<ll, ll>> HASH;
vector<pair<ll, ll>> p;

void build() {
    HASH.clear();
    HASH.resize(n + 1);
    pair<ll, ll> val = make_pair(0, 0);
    HASH[0] = val;
    for (int i = 0; i < n; i++) {
        val.first = (val.first * P1 + (int)s[i]) % MOD1;
        val.second = (val.second * P2 + (int)s[i]) % MOD2;
        HASH[i + 1] = val;
    }
    p.clear();
    p.resize(n + 1);
    val = make_pair(1, 1);
    p[0] = val;
    for (int i = 0; i < n; i++) {
        val.first = (val.first * P1) % MOD1;
        val.second = (val.second * P2) % MOD2;
        p[i + 1] = val;
    }
}

bool cmp(int i, int j, int len) {
    return ((HASH[i + len - 1].first - (HASH[i - 1].first * p[len].first) % MOD1 + MOD1) % MOD1 == (HASH[j + len - 1].first - (HASH[j - 1].first * p[len].first) % MOD1 + MOD1) % MOD1)
    && ((HASH[i + len - 1].second - (HASH[i - 1].second * p[len].second) % MOD2 + MOD2) % MOD2 == (HASH[j + len - 1].second - (HASH[j - 1].second * p[len].second) % MOD2 + MOD2) % MOD2);
}

void solve() {
    // 靠邀,我頭很痛。
    // 啊我微積分就退選,你覺得我看起來像會看懂題目的人嗎?
    cin >> s;
    n = s.size();
    build();
    int res = 0;
    string str = "";
    for (int len = 1; ; len++) {
        if (n / len < res) {
            break;
        }
        for (int i = 1; i + len <= n; i++) {
            for (int j = i + len; j + len - 1 <= n; j++) {
                if (cmp(i, j, len) && len > res) {
                    res = len;
                    str = s.substr(i - 1, len);
                }
            }
        }
    }
    cout << str << "\n";
    return ;
}

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

/*

5
TATCGATCGAGTTGT
TCCGCGAGCGAGTCTCTCCATT
GTTTCATCATACGAGGCCCCATACGCGCTGG
AGATGGGATCCTTATG
GCCCTTAGGCATGGGATGTCGTTTCTTG

*/

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3512kb

input:

5
TATCGATCGAGTTGT
TCCGCGAGCGAGTCTCTCCATT
GTTTCATCATACGAGGCCCCATACGCGCTGG
AGATGGGATCCTTATG
GCCCTTAGGCATGGGATGTCGTTTCTTG

output:

ATCG
GCGA
CATACG
GAT
CTT

result:

ok 5 lines

Test #2:

score: 0
Accepted
time: 5ms
memory: 3436kb

input:

50
TTGACAACTTCAGGTTGGCACTCCTTCATTTGGATTTCGGAATAATAGTTTTCTGCTCTGCC
ATCCTATTCGGGGATAGGAGAGATGGGTTGCCGCTATAAAAGCATTTGAACTCCATTTCACTCCGTTGGCTAGGGGTCGCACTG
CCGTAATATAAAGACTCGGAATTCCAATAGCTGCTATTTGCGAGTATGTGACTGAAAACACACCTATAAATATTAGCTGCGTACAAGCTA
ATGGCTGCATGCAGGGTCGACTAGACACACTTTGTCT
TTGAGGATGTCGACGTGTCT...

output:

CTTCA
CATTT
TAGCTGC
TGCA
ACGTG
GCGCCGG
CTCTT
AGTAT
AGAG
ACAG
TAT
TGAC
CTTG
CGTC
TACTGG
GCCGGT
GAA
CAGTA
GCGT
GGTT
CCCT
GAG
TAGAC
GGTGC
GCAGT
TGAG
ATCAA
CCACACA
GAGTC
ATGTA
ATGGTA
TATA
TATGAA
TTCC
CATACG
TACCA
TTAG
GGAATGT
CAGG
GCT
AAG
CTGT
GGAT
TCTTC
AAAAC
ATG
GATAA
TTA
ACATAT
CAAT

result:

ok 50 lines