# HackerEarth Path queries problem solution

In this HackerEarth Path queries problem solution, You are given an undirected tree G with N nodes. Every node is assigned a value denoted by .

A simple path u1 - u2 - u3 - u4,..., -uk is said to be beautiful if the number of pairs of nodes (ui, ui+1) where A[ui] is odd and A[ui + 1] is even and number of pairs of nodes (ui, ui+1) where A[ui] is even and A[ui + 1] is odd are equal.

Given Q queries of the form:

u val: Set the value of node n equal to val, i.e. set A[u] = val and find the number of ordered pairs (u,v) such that the simple path between node u and node v is beautiful.

## hackerEarth Path queries problem solution.

`#include<bits/stdc++.h>#define int long long intusing namespace std;void solve(){    int n, q;    cin >> n >> q;    assert(1 <= n and n <= 200000);    assert(1 <= q and q <= 200000);    int a[n + 1];    int c_e = 0;    int c_o = 0;    for(int i = 1 ; i <= n ; i++)    {        cin >> a[i];        assert(1 <= a[i] and a[i] <= 1000000000);        if(a[i]%2) c_o++;        else c_e++;    }    for(int i = 1 ; i <= n - 1 ; i++){        int u, v;        cin >> u >> v;        assert(1 <= u and u <= n);        assert(1 <= v and v <= n);        assert(u != v);    }    while(q--)    {            int u, val;            cin >> u >> val;            assert(1 <= u and u <= n);            assert(1 <= val and val <= 1000000000);            if(a[u]%2) c_o--;            else c_e--;            a[u] = val;            if(a[u]%2) c_o++;            else c_e++;                    int answer = (c_e*(c_e - 1))/2;            answer += (c_o*(c_o - 1))/2;            answer += (c_o + c_e);            cout << answer << " ";    }    cout << endl;}signed main(){    ios_base::sync_with_stdio(false);    cin.tie(NULL);    int t;    cin >> t;    assert(1 <= t and t <= 10);    while(t--){        solve();    }}`