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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#167842#7108. CouleurDelay_for_five_minutesTL 2ms5508kbC++144.4kb2023-09-07 17:48:112023-09-07 17:48:11

Judging History

你现在查看的是最新测评结果

  • [2023-09-07 17:48:11]
  • 评测
  • 测评结果:TL
  • 用时:2ms
  • 内存:5508kb
  • [2023-09-07 17:48:11]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
int a[100005];
int B ;
ll sum[325][325] ;
int be[100005] , L[325] , R[325];
vector<int> va[325];
struct BIT {
    int c[100005];
    void reset() {
        for(int i = 1;i <= n;i++) c[i] = 0;
    }
    void add(int u,int v) {
        while(u <= n) {
            c[u] += v;u += (u & -u) ;
        }
    }
    int qry(int u) {
        int ans = 0;
        while(u) {
            ans += c[u] ; u -= (u & -u);
        }
        return ans;
    }
}fen ;
int cal_inv(int l,int r)
{
    int ans = 0;
    for(int i = r;i >= l;i--) {
        ans += fen.qry(a[i] - 1) ;
        fen.add(a[i] , 1) ;
    }
    for(int i = l;i <= r;i++) fen.add(a[i], -1);
    return ans;
}
vector<int> get(int l,int r)
{
    vector<int> ans;
    for(int i = l;i <= r;i++) ans.push_back(a[i]) ;
    return ans;
}
ll gb(vector<int>& a,vector<int> &b) ///a[i] > b[j];
{
    int np = 0;
    ll ans = 0;
    for(int i = 0;i < a.size();i++) {
        while(np < b.size() && b[np] < a[i]) np++;
        ans += np;
    }
    return ans;
}
ll sums[325][100005];
void init()
{
    B = sqrt(n) ;
    for(int i = 1;i <= n;i++) {
        be[i] = be[i - 1];
        if(i % B == 1 || B == 1) {be[i]++; L[be[i]] = i ; R[be[i - 1]] = i - 1;}
    }
    R[be[n]] = n;
    for(int i = 1;i <= be[n];i++) {va[i] = get(L[i] , R[i]) ; sort(va[i].begin() , va[i].end()) ;}
    for(int i = 1;i <= be[n];i++) {
        for(int j = i;j <= be[n];j++) {
            if(i == j) sum[i][j] = cal_inv(L[i] , R[i]) ;
            else {
                sum[i][j] = gb(va[i] , va[j]) ;
            }
        }
    }
    for(int i = 1;i <= be[n];i++) {
        for(int j = i + 1;j <= be[n];j++) sum[i][j] += sum[i][j - 1];
    }
    for(int j = be[n] ; j >= 1;j--) {
        for(int i = j - 1;i >= 1;i--) sum[i][j] += sum[i + 1][j] ;
    }
    for(int i = 1;i <= be[n];i++) {
        for(int j = 1;j <= n;j++) {
            if(L[i] <= j && j <= R[i]) sums[i][j] = sums[i][j - 1];
            else if(j < L[i]) {
                sums[i][j] = sums[i][j - 1] + (lower_bound(va[i].begin() , va[i].end() , a[j]) - va[i].begin()) ;
            }
            else {
                sums[i][j] = sums[i][j - 1] + (va[i].size() - (upper_bound(va[i].begin() , va[i].end() , a[j]) - va[i].begin())) ;
            }
        }
    }
}
ll qry(int l,int r)
{
    if(be[l] == be[r]) {
        return cal_inv(l , r) ;
    }
    int Lp = be[l] , Rp = be[r] ;
    ll ans = sum[Lp + 1][Rp - 1];
  //  printf("B %lld\n" , ans);
    for(int i = Lp +1; i <= Rp-1;i++) {
        ans += sums[i][R[Lp]] - sums[i][l - 1];
        ans += sums[i][r] - sums[i][L[Rp] - 1] ;
    }
    vector<int> x = get(l , R[Lp]) , y = get(L[Rp] , r) ;
    sort(x.begin(),x.end()) ; sort(y.begin() , y.end()) ;
    ans += gb(x , y);
      //  printf("After %lld\n" , ans);
    ans += cal_inv(l , R[Lp]) + cal_inv(L[Rp] , r) ;
    return ans;
}
struct val {
    int l , r;
    ll v;
};
bool operator < (val x,val y)
{
    return x.r < y.r;
}
void solv()
{
    set<val> st;
    multiset<ll> vals;
    cin>>n;
    for(int i=1;i<=n;++i)cin>>a[i];
    init();
    ll z = qry(1 , n) ; cout << z <<' ' ;
    st.insert((val){1 , n , z}) ; vals.insert(z) ;
    // puts("ojbk") ; printf("Z %lld\n",z) ;
    // exit(0) ;
    // printf("ASK %lld\n",qry(2 , n)) ;
    for(int i = 1;i <= n;i++) {
        long long x ; cin >> x;
        x ^= z;
        auto it = st.lower_bound((val){x , x , 0}) ;
        int ul = it->l , ur = it->r;
        ll uv = it->v ;

        vals.erase(vals.find(uv)) ;
        st.erase(it) ;
        //printf("ul %d , ur %d\n",ul,ur) ;
        if(x - 1 >= ul) {
            ll nv = qry(ul , x - 1) ; //printf("Lnv %lld\n",nv);
            st.insert((val){ul , x - 1 , nv}) ; vals.insert(nv) ;
        }
        if(x + 1 <= ur) {
            ll nv = qry(x + 1 , ur); //printf("Rnv %lld\n",nv);
            st.insert((val){x + 1 , ur , nv}) ; vals.insert(nv) ;
        }


        z = 0;
        if(vals.size()) {
            auto it = vals.end() ; it-- ; z = (*it) ;
        }
        if(i != n) cout << z <<' ' ;
        //printf("Z %lld\n",z) ;
    }
    cout << '\n' ;
}
int main(){
    ios::sync_with_stdio(false) ; cin.tie(0) ; cout.tie(0) ;
  //  freopen("in.txt","r",stdin) ;
    int t;cin >> t;
    while(t--) solv() ;
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 2ms
memory: 5508kb

input:

3
5
4 3 1 1 1
5 4 5 3 1
10
9 7 1 4 7 8 5 7 4 8
21 8 15 5 9 2 4 5 10 6
15
4 8 8 1 12 1 10 14 7 14 2 9 13 10 3
37 19 23 15 7 2 10 15 2 13 4 5 8 7 10

output:

7 0 0 0 0 
20 11 7 2 0 0 0 0 0 0 
42 31 21 14 14 4 1 1 1 0 0 0 0 0 0 

result:

ok 3 lines

Test #2:

score: -100
Time Limit Exceeded

input:

11116
10
10 5 10 3 6 4 8 5 9 8
31 27 24 11 12 3 0 2 3 1
10
8 2 7 2 8 10 1 10 9 10
6 5 2 13 2 1 0 1 3 1
10
7 10 7 6 1 3 10 6 7 9
21 18 10 1 6 5 4 8 9 10
10
2 10 4 8 8 5 7 2 6 7
20 10 9 1 15 0 4 2 9 7
10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
10
1 2 3 4 5 6 7 8 9 10
6 3 5 2 7 10 9 1 4 8
10
1 10 1 3...

output:

21 18 16 12 10 6 4 1 1 0 
12 12 10 10 4 4 4 2 1 0 
20 16 9 5 3 3 3 0 0 0 
22 14 8 8 5 5 2 1 1 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
19 12 7 4 4 2 2 1 0 0 
20 18 8 3 1 1 0 0 0 0 
45 21 21 10 3 3 3 0 0 0 
17 11 8 2 1 1 1 0 0 0 
13 4 1 0 0 0 0 0 0 0 
29 27 22 15 9 7 4 3 1 0 
26 16 9 2 1 1 1 1 1 ...

result: