제출 #58713

#제출 시각아이디문제언어결과실행 시간메모리
58713Benq원형 문자열 (IZhO13_rowords)C++14
32 / 100
2072 ms424 KiB
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

string a,b;
array<int,2001> cur;

int lcs() {
    cur.fill(0);
    for (char c: b) {
        FORd(i,1,sz(a)+1) if (a[i-1] == c) cur[i] = cur[i-1]+1;
        FOR(i,1,sz(a)+1) cur[i] = max(cur[i],cur[i-1]);
    }
    return cur[sz(a)];
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> a >> b;
    int ans = 0;
    F0R(i,sz(a)) {
        ans = max(ans,lcs());
        rotate(a.begin(),a.begin()+1,a.end());
    }
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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