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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#488927#7956. Walk Swappingarnold518WA 0ms3960kbC++174.1kb2024-07-24 16:23:302024-07-24 16:23:30

Judging History

This is the latest submission verdict.

  • [2024-07-24 16:23:30]
  • Judged
  • Verdict: WA
  • Time: 0ms
  • Memory: 3960kb
  • [2024-07-24 16:23:30]
  • Submitted

answer

#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int MAXN = 3000;
const int INF = 1e9;

template<int MOD>
struct mint
{
    int x;
    mint() : x(0) {}
    mint(int _x) : x(_x) {}
    mint operator + (int ot) const { return x+ot>=MOD ? x+ot-MOD : x+ot; }
    mint operator - (int ot) const { return x<ot ? x+MOD-ot : x-ot; }
    mint operator - () const { return x ? MOD-x : 0; }
    mint operator * (int ot) const { return 1ll*x*ot%MOD; }
    mint operator += (int ot) { return *this = *this + ot; }
    mint operator -= (int ot) { return *this = *this - ot; }
    mint operator *= (int ot) { return *this = *this * ot; }
    operator int() const { return x; }
    bool operator < (int ot) const { return x<ot; }
};

template<int MOD>
mint<MOD> mpow(mint<MOD> a, ll b)
{
    mint<MOD> ret=1;
    while(b)
    {
        if(b&1) ret*=a;
        b>>=1; a=a*a;
    }
    return ret;
}
template<int MOD>
mint<MOD> inv(mint<MOD> a) { return mpow<MOD>(a, MOD-2); }

const int MOD1 = 1e9+7;
const int MOD2 = 998244353;
const int P1 = 3;
const int P2 = 7;

typedef pair<mint<MOD1>, mint<MOD2>> pmm;

pmm operator + (const pmm &p, const pmm &q) { return {p.first+q.first, p.second+q.second}; }
pmm operator - (const pmm &p, const pmm &q) { return {p.first-q.first, p.second-q.second}; }
pmm operator * (const pmm &p, const pmm &q) { return {p.first*q.first, p.second*q.second}; }
pmm operator - (const pmm &p) { return {-p.first, -p.second}; }
pmm& operator += (pmm &p, const pmm &q) { p=p+q; return p; }
pmm& operator -= (pmm &p, const pmm &q) { p=p-q; return p; }
pmm& operator *= (pmm &p, const pmm &q) { p=p*q; return p; }
pmm mpow(pmm a, ll b) { return {mpow(a.first, b), mpow(a.second, b)}; }
pmm inv(pmm a) { return {inv(a.first), inv(a.second)}; }

int N, A[MAXN+10], B[MAXN+10];
mint<MOD1> PP[MAXN+10], IP[MAXN+10];
mint<MOD1> HA[MAXN+10], HB[MAXN+10];
int ans=INF;

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    
    scanf("%d", &N);
    for(int i=1; i<=N; i++) scanf("%d", &A[i]);
    for(int i=1; i<=N; i++) scanf("%d", &B[i]);
    
    {
        vector<int> VA, VB;
        for(int i=1; i<=N; i++) VA.push_back(A[i]);
        for(int i=1; i<=N; i++) VB.push_back(B[i]);
        sort(VA.begin(), VA.end());
        sort(VB.begin(), VB.end());
        if(VA!=VB) return !printf("-1\n");
    }

    if(N==1) return !printf("0\n");

    PP[0]=1; IP[0]=1;
    PP[1]=P1; IP[1]=inv(PP[1]);
    for(int i=2; i<=N; i++) PP[i]=PP[i-1]*PP[1];
    for(int i=2; i<=N; i++) IP[i]=IP[i-1]*IP[1];

    for(int cyc=0; cyc<N; cyc++)
    {
        vector<pii> V, V2, V3;
        for(int i=1; i<=N; i++) HA[i]=HA[i-1]+PP[i]*A[i];
        for(int i=1; i<=N; i++) HB[i]=HB[i-1]+PP[i]*B[i];

        for(int i=1; i<=N; i++) V.push_back({HB[i-1]+(HB[N]-HB[i])*IP[1], i});
        for(int i=2; i<=N; i++) V2.push_back({(HA[N]-HA[i-1])*IP[i-1] + (HA[i-1]-HA[1])*PP[N-i], i});
        for(int i=1; i<N; i++) V3.push_back({(HA[N-1]-HA[i-1])*IP[i-1] + HA[i-1]*PP[N-i], i});

        sort(V.begin(), V.end());
        sort(V2.begin(), V2.end());
        sort(V3.begin(), V3.end());

        for(int l=0, r, j=0; l<V2.size(); l=r)
        {
            for(r=l; r<V2.size() && V2[r].first==V2[l].first; r++);
            int val=INF;
            for(; j<V.size() && V[j].first<=V2[l].first; j++) if(V[j].first==V2[l].first) val=min(val, V[j].second);
            if(val!=INF) for(int i=l; i<r; i++) ans=min(ans, (V2[i].second-2)*N+val-1);
        }
        for(int l=0, r, j=0; l<V3.size(); l=r)
        {
            for(r=l; r<V3.size() && V3[r].first==V3[l].first; r++);
            int val=0;
            for(; j<V.size() && V[j].first<=V3[l].first; j++) if(V[j].first==V3[l].first) val=max(val, V[j].second);
            if(val!=0) for(int i=l; i<r; i++) ans=min(ans, (N-V3[i].second)%(N-1)*N+N-val);
        }

        rotate(A+1, A+2, A+N+1);
        rotate(B+1, B+2, B+N+1);
    }

    if(ans==INF) printf("-1\n");
    else printf("%d\n", ans);
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3868kb

input:

4
4 3 2 1
3 4 2 1

output:

1

result:

ok single line: '1'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3960kb

input:

6
2 1 1 2 2 1
1 2 2 2 1 1

output:

7

result:

ok single line: '7'

Test #3:

score: 0
Accepted
time: 0ms
memory: 3880kb

input:

6
4 1 3 6 2 5
6 2 1 3 4 5

output:

-1

result:

ok single line: '-1'

Test #4:

score: 0
Accepted
time: 0ms
memory: 3860kb

input:

4
1 2 3 4
4 2 1 3

output:

2

result:

ok single line: '2'

Test #5:

score: -100
Wrong Answer
time: 0ms
memory: 3952kb

input:

6
4 1 3 6 2 5
6 2 5 3 4 1

output:

9

result:

wrong answer 1st lines differ - expected: '13', found: '9'