Kill Demodogs

题意

大意是求

思路

n项平方和求法：
1 + 4 + 9 + 16 + .... + $n^2$
= 1 * 1 + 2 * 2 + 3 * 3 + ... + n * n
= (1 + 2 + 3 + ... + n) + (1 * 2 + 2 * 3 + ... + (n - 1) * n)
= n * (n + 1）/ 2 + (1 * 2 * 3 + 2 * 3 * 3 + 3 * 3 * 4 + ... + 3 * (n - 1) * n) / 3
= n * (n + 1) / 2 + (1 * 2 * 3 + 2 * 3 * 4 - 1 * 2 * 3 + 3 * 4 * 5 - 2 * 3 * 4 + ... + (n - 1) * n * (n + 1)) / 3
= n * (n + 1) / 2 + (n - 1) * n * (n + 1) / 3

(1 * 2 + 2 * 3 + ... + (n - 1) * n)
= (1 * 2 * 3 + 2 * 3 * 3 + 3 * 3 * 4 + ... + 3 * (n - 1) * n) / 3
= (1 * 2 * 3 + 2 * 3 * 4 - 1 * 2 * 3 + 3 * 4 * 5 - 2 * 3 * 4 + ... + (n - 1) * n * (n + 1)) / 3
= (n - 1) * n * (n + 1) / 3

相加即可

void solve() {
	LL n;
	cin >> n;
	LL sum = 0;
	if (n % 3 == 0) {
		sum = n * (n + 1) / 2 % MOD + n * (n + 1) / 3 % MOD * (n - 1) % MOD + n * (n + 1) / 3 % MOD * (n - 1) % MOD;
	}else {
		sum = n * (n + 1) / 2 % MOD + (n - 1) * (n + 1) / 3 % MOD * n % MOD + (n - 1) * (n + 1) / 3 % MOD * n % MOD;
	}
	sum = 2022 * sum % MOD;
	cout << sum << endl;
}