#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
pair<int, int> s[110];
int d[110];
void get() {
	int n, l = 0, i, c;
	cin >> n;
	for (i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			c = 0;
			while (n % i == 0) c++, n /= i;
			s[++l] = make_pair(c, i);
		}
	}
	if (n != 1) s[++l] = make_pair(1, n);
	sort(s + 1, s + l + 1), d[l + 1] = 1;
	for (i = l; i >= 1; i--) d[i] = d[i + 1] * s[i].second;
	int ans = 0;
	for (i = 1; i <= l; i++) if (s[i].first != s[i - 1].first) ans += d[i] * (s[i].first - s[i - 1].first);
	cout << ans << endl;
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T;
	cin >> T;
	while (T--) get();
	return 0;
}

This is the reason why the problem was named as Remove the Bracket.

However, We discussed to remove it on 28th Jan in the statement. Really sorry for inconvenience of the statement!

#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long f[N][2],x[N],y[N];
void get() {
	int i,n,s,j;
	cin>>n>>s;
	for(i=1; i<=n; i++) {
		cin>>j;
		if(i==1||i==n) x[i]=y[i]=j;
		else if(j<=s) x[i]=0,y[i]=j;
		else x[i]=s,y[i]=j-s;
	}
	f[1][0]=f[1][1]=0;
	for(i=2; i<=n; i++) {
		f[i][0]=min(f[i-1][0]+y[i-1]*x[i],f[i-1][1]+x[i-1]*x[i]);
		f[i][1]=min(f[i-1][0]+y[i-1]*y[i],f[i-1][1]+x[i-1]*y[i]);
	}
	cout<<f[n][0]<<endl;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T; cin>>T;
	while(T--) get();
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int a[N];
int v[N]; //= 1 -> in the tree with the end node
int s[N]; //subtree size
struct E {
	int to;
	E *nex;
} *h[N];
void add(int u, int v) {
	E *cur = new E;
	cur->to = v, cur->nex = h[u], h[u] = cur;
}
void dfs(int u) {
	s[u] = v[u] = 1;
	for (E *cur = h[u]; cur; cur = cur->nex)
		dfs(cur->to), s[u] += s[cur->to];
}
void get() {
	int i, j, n;
	cin >> n;
	for (i = 1; i <= n + 1; i++)
		s[i] = v[i] = 0, h[i] = 0;
	for (i = 1; i <= n; i++) {
		cin >> a[i], a[i] = min(i + a[i], n + 1);
		if (a[i] <= 0) a[i] = n + 1;
		add(a[i], i);
	}
	dfs(n + 1); //start with the end point, dfs the tree
	long long ans = 0;
	if (v[1] == 1) {
		j = 1; do { ans -= s[j] + (n - s[n + 1] + 1), j = a[j]; }
		while (j != n + 1);
		ans += 1ll * n * (2 * n + 1);
	} else {
		j = 1; do { ans += (n + s[n + 1]), v[j] = 2, j = a[j]; }
		while (v[j] != 2);
	}
	cout << ans << endl;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T;
	cin >> T;
	while (T--) get();
	return 0;
}

Bonus Problem: What if $$$x=0$$$?

#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int dest[N], visit[N], ans[N], mp[N], ansinv[N], sp;
int pow2(int y, int M) {
	long long x = 2, ans = 1;
	while (y) {
		if (y & 1) ans = ans * x % M;
		x = x * x % M, y >>= 1;
	}
	return ans % M;
}
vector<int> cyc[N];
void get() {
	sp = 0;
	int n, i, j, m, z, o;
	cin >> n >> m;
	for (i = 1; i <= n; i++) cyc[i].clear();
	for (i = 1; i <= n; i++) visit[i] = 0, cin >> dest[i];
	for (i = 1; i <= n; i++) {
		if (visit[i]) continue;
		j = i, z = 0; do { ++z, visit[j] = 1, j = dest[j]; } while (j != i);
		cyc[z].push_back(i);
	}
	int lim = pow2(min(m, 20), 1e9), s = 0, d, t, num, cp;
	for (i = 1; i <= n; i++) {
		if (!cyc[i].size()) continue;
		int siz = cyc[i].size();
		if (i % 2 == 0 && siz % lim) {
			cout << "NO" << endl; return;
		} else {
			cp = -1;
			for (j = lim; j > 0; j /= 2) {
				while (siz >= j) {
					siz -= j, d = s + j * i, t = j;
					while (t--) {
						++s, ++cp, num = s, o = cyc[i][cp];
						do {
						    ans[num] = o, num += pow2(m, i * j);
						    if (num > d) num -= j * i;
						    o = dest[o];
						} while (o != cyc[i][cp]);
					}
					for (int k = sp + 1; k < sp + j * i; k++) mp[k] = k + 1;
					mp[sp + j * i] = sp + 1, sp += j * i, s = d;
				}
			}
		}
	}
	cout << "YES" << endl;
	for (i = 1; i <= n; i++) ansinv[ans[i]] = i;
	for (i = 1; i <= n; i++) cout << ans[mp[ansinv[i]]] << " \n"[i == n];
}
int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	int T;
	cin >> T;
	while (T--) get();
	return 0;
}

UPD: An alternative implementation using segment tree comment link.

Have to declare that we built up Problem H from zero. ;)