HackerEarth Alice's construction problem solution

In this HackerEarth Alice’s construction problem solution Alice has a warehouse with n boxes and the ith box has number pi. All box numbers are different and range from 1 to n. Alice wants to rearrange the boxes in the following way:

First, she chooses some interval of boxes, starting from the Lth to the Rth inclusive, as well as some number X.

Then arbitrarily chooses a pair of indices i and j from the interval [L;R] such that |i – j| = 1 and pi <= X, pj <= X and changes the boxes at positions i and j. That is, an arbitrary number of times swaps a pair of adjacent boxes in which the numbers do not exceed X.

You are given m triplets Li, Ri, Xi. For each of them, Alice asks you to tell how many different sequences of boxes she can get by selecting the following values.

Since these numbers can be very large, you need to subtract the remainder of their division by the prime number 10^9 + 7.

HackerEarth Alice’s construction problem solution.

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

const int MAX_N = 1e5 + 10;

struct my{
    int cntPref,cntSuff;
    int cntAll;
    int val;
    my(){
     cntPref = 0;
     cntSuff = 0;
     cntAll = 1;
     val = 1;
    }
};

int f[MAX_N];
my t[4 * MAX_N];
int md = 1e9 + 7;
int p[MAX_N],pos[MAX_N];
vector<pair<pair<int,int>,int>> q[MAX_N];
int answer[MAX_N];
int rf[MAX_N];

int BinPow(int a,int b){
    int ans = 1;
    while(b){
        if(b & 1)ans = ans * a % md;
        a = a * a % md;
        b >>= 1;
    }
    return ans;
}

my combine(my a,my b){
    my c = my();
    c.cntAll = a.cntAll + b.cntAll;
    if(b.cntPref > 0 && a.cntSuff > 0){
        if(a.cntAll == a.cntSuff && b.cntAll == b.cntPref){
            c.cntPref = c.cntAll;
            c.cntSuff = c.cntAll;
            c.val = a.val * b.val % md * rf[a.cntSuff]%md * rf[b.cntPref]%md*f[a.cntSuff + b.cntPref] % md;
        }else
        if(a.cntAll == a.cntSuff){
            c.cntPref = a.cntAll + b.cntPref;
            c.cntSuff = b.cntSuff;
            c.val = a.val * b.val % md * rf[a.cntSuff]%md * rf[b.cntPref]%md*f[a.cntSuff + b.cntPref] % md;
        }else
        if(b.cntAll == b.cntPref){
            c.cntPref = a.cntPref;
            c.cntSuff = b.cntAll + a.cntSuff;
            c.val = a.val * b.val % md * rf[a.cntSuff]%md * rf[b.cntPref]%md*f[a.cntSuff + b.cntPref] % md;
        }else{
            c.cntPref = a.cntPref;
            c.cntSuff = b.cntSuff;
            c.val = a.val * b.val % md * rf[a.cntSuff]%md * rf[b.cntPref]%md*f[a.cntSuff + b.cntPref] % md;
        }
    }else{
        c.cntPref = a.cntPref;
        c.cntSuff = b.cntSuff;
        c.val = (a.val * b.val) % md;
    }
    return c;
}

void upd(int v,int tl,int tr,int pos){
    if(tl > pos || tr < pos)return;
    if(tl == tr){
        t[v].cntPref = t[v].cntSuff = t[v].cntAll = t[v].val = 1;
        return;
    }
    int tm = (tl + tr) >> 1,
        L = v << 1,
        R = L | 1;
    upd(L, tl, tm, pos);
    upd(R, tm + 1, tr, pos);
    t[v] = combine(t[L], t[R]);
}

my get(int v,int tl,int tr,int l,int r){
    if(tl >= l && tr <= r){
        return t[v];
    }
    int tm = (tl + tr) >> 1,
        L = v << 1,
        R = L | 1;
    if(r <= tm)return get(L, tl, tm, l, r);
    if(l > tm)return get(R, tm + 1, tr, l, r);
    my a1 = get(L, tl, tm, l, r),
        a2 = get(R, tm + 1, tr, l, r);
    my a3 = combine(a1, a2);
        return combine(a1, a2);
}

 main(){

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int n, m;
    cin >> n >> m;

    f[0] = 1;
    rf[0] = 1;
    for(int i = 1; i <= n; ++i){
        f[i] = f[i - 1] * i % md;
        rf[i] = BinPow(f[i], md - 2);
    }

    for(int i = 1; i <= 4 * n; ++i)
        t[i] = my();

    for(int i = 1; i <= n; ++i){
        cin >> p[i];
        pos[p[i]] = i;
    }

    for(int i = 1; i <= m; ++i){
        int l, r, x;
        cin >> l >> r >> x;
        q[x].push_back({{l, r},i});
    }

    for(int i = 1; i <= n; ++i){
        upd(1, 1, n, pos[i]);
        for(int j = 0; j < (int)q[i].size(); ++j){
            answer[q[i][j].second] = get(1, 1, n, q[i][j].first.first, q[i][j].first.second).val;
        }
    }

    for(int i = 1; i <= m; ++i)
        cout << answer[i] << 'n';

    return 0;
}

Second solution

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

const int MAX_N = 1e5 + 14, MOD = 1e9 + 7;

int n, q, p[MAX_N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n >> q;
    for (int i = 0; i < n; ++i)
        cin >> p[i];
    for (int i = 0; i < q; ++i) {
        int l, r, x;
        cin >> l >> r >> x;
        --l;
        int ans = 1;
        for (int j = l, c = 0; j < r; ++j)
            if (p[j] <= x)
                ans = (ll) ans * ++c % MOD;
            else
                c = 0;
        cout << ans << 'n';
    }
}

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