#include <iostream>
#include <vector>

using namespace std;

int main()
{
	int tests;
	cin >> tests;
	while (tests--) {
		int n;
		cin >> n;

		vector<int> cnt(26);
		for (int i = 0; i < n; ++i) {
			string s;
			cin >> s;
			for (char ch : s) {
				++cnt[ch - 'a'];
			}
		}

		bool ans = true;
		for (int i = 0; i < 26; ++i) {
			if (cnt[i] % n != 0) {
				ans = false;
				break;
			}
		}

		cout << (ans ? "YES" : "NO") << endl;
	}
}

numTests = int(input())
for testNo in range(numTests):
	n = int(input())
	cnt = [0 for i in range(26)]
	for _ in range(n):
	    s = input()
	    for i in s:
	        cnt[ord(i) - 97] += 1

	ans = True
	for i in range(26):
	    if cnt[i] % n != 0:
	        ans = False
	        break

	if ans:
	    print('YES')
	else:
	    print('NO')

The cost to transform $$$a_i$$$ to $$$c^i$$$ is $$$\lvert a_i - c^i \rvert$$$, and $$$\lvert a_i - c^i \rvert + \lvert a_j - c^j \rvert \le \lvert a_j - c^i \rvert + \lvert a_i - c^j \rvert$$$ when $$$i < j$$$ and $$$a_i \le a_j$$$, thus it is optimal to have $$$a_i \le a_j$$$ for each $$$0 \le i < j < n$$$.

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>

using namespace std;

const int64_t INF = 1e17;
inline int64_t mul(int64_t a, int64_t b)
{
	return (INF / a > b ? a * b : INF);
}

inline int64_t add(int64_t a, int64_t b)
{
	return (a + b >= INF ? INF : a + b);
}

int main()
{
	int n;
	cin >> n;

	vector<int> a(n);
	for (int &x : a) cin >> x;
	sort(a.begin(), a.end());

	if (n <= 2) {
		cout << a[0] - 1 << endl;
	} else {
		int64_t ans = accumulate(a.begin(), a.end(), 0ll) - n;

		for (int x = 1; ; ++x) {
			int64_t curPow = 1, curCost = 0;
			for (int i = 0; i < n; ++i) {
				curCost = add(curCost, abs(a[i] - curPow));
				curPow = mul(curPow, x);
			}

			if (curPow == INF || curPow / x > ans + a[n - 1]) break;
			ans = min(ans, curCost);
		}

		cout << ans << endl;
	}
}

n = int(input())
a = [int(x) for x in input().split()]
a.sort()
inf = 10**18

if n <= 2:
    print(a[0] - 1)
else:
    ans = sum(a) - n

    for x in range(1, 10**9):
        curPow = 1
        curCost = 0
        for i in range(n):
            curCost += abs(a[i] - curPow)
            curPow *= x
            if curPow > inf:
                break

        if curPow > inf:
            break
        if curPow / x > ans + a[n - 1]:
            break

        ans = min(ans, curCost)

    print(ans)

$$$1 \space \space 1$$$
$$$0$$$
$$$1 \space \space 1$$$
$$$0$$$
$$$1 \space \space 1$$$
$$$-a_1$$$

$$$1 \space \space 1$$$
$$$-a_1$$$
$$$1 \space \space n$$$
$$$0 \space \space -n \cdot a_2 \space \space -n \cdot a_3 \dots -n \cdot a_n$$$
$$$2 \space \space n$$$
$$$(n-1) \cdot a_2 \space \space (n-1) \cdot a_3 \space \space \dots \space \space (n-1) \cdot a_n$$$

Let us prove by induction:

When $$$S = 0$$$, the second player obviously wins.

When $$$S \geq 2$$$, consider the game state after the first player moves. If there is a pile that now has more than $$$\lfloor \frac{S}{2} \rfloor$$$ stones, then we arrive back at case A where the next player to move wins. Otherwise, the second player can choose from any valid pile (note that the case condition implies that there are at least two non-empty piles before the first player's move). Now $$$S$$$ has been reduced by $$$2$$$, and every pile still has at most $$$\lfloor \frac{S}{2} \rfloor$$$ stones.

The condition allows us to assign a perfect matching of stones, where one stone is matched with exactly one stone from a different pile.

A greedy way to create such a matching: Give each label $$$0, 1, \dots, S - 1$$$ to a different stone so that for every pair of stones with labels $$$l < r$$$ that are from the same pile, stones $$$l + 1, l + 2, \dots, r - 1$$$ are also from that pile; then match stones $$$i$$$ with $$$i + \frac{S}{2}$$$ for all $$$0 \le i < \frac{S}{2}$$$.

For every stone that the first player removes, the second player can always remove its matching stone, until the first player can no longer make a move and loses.

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>

using namespace std;

int main()
{
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;

		vector<int> a(n);
		for (int &x : a) cin >> x;

		int maxPile = *max_element(a.begin(), a.end());
		int numStones = accumulate(a.begin(), a.end(), 0);

		if (maxPile * 2 > numStones || (numStones & 1)) cout << "T" << endl;
		else cout << "HL" << endl;
	}
}

t = int(input())
for _ in range(t):
    n = int(input())
    a = [int(x) for x in input().split()]

    maxPile = max(a)
    numStones = sum(a)

    if maxPile * 2 > numStones or (numStones & 1):
        print('T')
    else:
        print('HL')

/*input
4 2 4 4 1
4 5 1 2
*/
#include <bits/stdc++.h>
using namespace std;

int read() {
	int x = 0, c = getchar();
	for(; !(c > 47 && c < 58); c = getchar());
	for(; (c > 47 && c < 58); c = getchar()) x = x * 10 + c - 48;
	return x;
}

void upd(long long &a, long long b) {
	a = (a < b) ? a : b;
}

const int N = 1e6 + 5;

long long f[N][2];
int n, r1, r2, r3, d, a[N];

int main(){ 
	n = read(), r1 = read(), r2 = read(), r3 = read(), d = read();
	for(int i = 1; i <= n; a[i ++] = read());

	for(int i = 2; i <= n; ++ i) f[i][0] = f[i][1] = 1e18;

	f[1][0] = 1ll * r1 * a[1] + r3;
	f[1][1] = min(0ll + r2, 1ll * r1 * a[1] + r1);
	for(int i = 1; i < n; ++ i) {
		// 0 -> 0
		// so we clear this one and the next one as well
		upd(f[i + 1][0], f[i][0] + d + 1ll * r1 * a[i + 1] + r3);

		// 0 -> 1
		// this one is cleared, but next one isnt
		upd(f[i + 1][1], f[i][0] + d + min(0ll + r2, 1ll * r1 * a[i + 1] + r1));
		
		// 1 -> 0
		upd(f[i + 1][0], f[i][1] + d + 1ll * r1 * a[i + 1] + r3 + 2 * d + r1);
		upd(f[i + 1][0], f[i][1] + d + 1ll * r1 * a[i + 1] + r1 + d + r1 + d + r1);
		upd(f[i + 1][0], f[i][1] + d + r2 + d + r1 + d + r1);

		// 1 -> 1
		upd(f[i + 1][1], f[i][1] + d + r2 + d + r1 + d);
		upd(f[i + 1][1], f[i][1] + d + 1ll * r1 * a[i + 1] + r1 + d + r1 + d);

		if(i == n - 1) {
			upd(f[i + 1][0], f[i][1] + d + 1ll * r1 * a[i + 1] + r3 + d + r1);
		}
	}
	cout << f[n][0] << endl;
}