제출 #1119298

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1119298		MateiKing80	Sum Zero (RMI20_sumzero)	C++14	61 / 100	233 ms	24948 KiB

sumzero

#include <bits/stdc++.h>

using namespace std;

int tata[1 << 19], ptr[1 << 19];
int a[1 << 19], ans[1 << 19];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++)
        cin >> a[i], a[i] += a[i - 1];
    unordered_map<int, int> mp;
    int last = n + 1;
    mp[a[n]] = n;
    ans[n + 2] = -1;
    ans[n + 1] = 0;
    tata[n + 1] = n + 2;
    tata[n + 2] = n + 2;
    for(int i = n; i; i --)
    {
        if(mp[a[i - 1]] != 0)
            last = min(last, mp[a[i - 1]]);
        mp[a[i - 1]] = i - 1;
        tata[i] = last + 1;
        ans[i] = 1 + ans[last + 1];
    }
    int q;
    cin >> q;
    for(int i = n + 2; i; i --)
    {
        if(ans[ptr[ptr[tata[i]]]] - ans[ptr[tata[i]]] == ans[ptr[tata[i]]] - ans[tata[i]])
            ptr[i] = ptr[ptr[tata[i]]];
        else
            ptr[i] = tata[i];
    }
    for(int i = 1; i <= q; i ++)
    {
        int l, r, li;
        cin >> l >> r;
        li = l;
        r ++;
        while(1)
        {
            if(ptr[l] <= r)
                l = ptr[l];
            else if(tata[l] <= r)
                l = tata[l];
            else
                break;
        }
        cout << ans[li] - ans[l] << '\n';
    }
}
/*
ne trebuie prima pereche de sume partiale egale care se afla
la dreapta unui punct
si facem binary lifting pe tot
yessir NU AM MEMORIE CE E ASTA
am o idee
putem adauga incet incet lasturi de la dreapta la stanga
putem sa facem binary lifting in O(1) memorie ca la arbori
si adancimea va fi practic raspunsu?
ca e cazul ala
if(adanc[ptr[ptr[tata]]] - adanc[ptr[tata]] == adanc[ptr[tata]] - adanc[tata])
    ptr[nod] = ptr[ptr[tata]];
si cam asta e tot sper
*/
/*
10
1 2 -3 0 1 -4 3 2 -1 1
3
1 10
1 5
2 9

*/

• Subtask 1: 22 / 22
• Subtask 2: 39 / 39
• Subtask 3: 0 / 39