题意

$n$ 个数，每个数 $a_i$ ，选 $k$ 个数，一个选取方案的值是异或和。求所有选取方案值的和。

$n\leq10^5,a_i<2^{31}$ 。

题解

加法和异或本质相同。

考虑每一位的贡献，简单组合计数。

代码

const int N = 3e5 + 5, mod = 998244353;

inline ll Read() {
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

namespace Main {
	int n, m;
	ll a[N], frac[N], inv[N], ans;
	ll qpow (ll a, ll b) {
		ll ans = 1;
		for (; b; b >>= 1, a = 1ll * a * a % mod) 
			if (b & 1) ans = 1ll * ans * a % mod;
		return ans;
	}
	ll C (ll n, ll m) {
		if (m > n) return 0;
		return 1ll * frac[n] * inv[n - m] % mod * inv[m] % mod;
	}
	int main () {
		n = Read(), m = Read();
		for (int i = 1; i <= n; i++) a[i] = Read();
		frac[0] = 1;
		for (int i = 1; i <= n; i++) frac[i] = 1ll * frac[i - 1] * i % mod;
		inv[n] = qpow(frac[n], mod - 2);
		for (int i = n; i; i--) inv[i - 1] = 1ll * inv[i] * i % mod;
		for (int j = 0; j < 31; j++) {
			int cnt = 0;
			for (int i = 1; i <= n; i++) cnt += (a[i] >> j) & 1;
			for (int i = max(m - (n - cnt), 0); i <= min(cnt, m); i++)
				if (i & 1) ans = (ans + (1ll * C(cnt, i) * C(n - cnt, m - i) % mod) * qpow(2, j) % mod) % mod;
		}
		printf ("%lld\n", ans);
		return 0;
	}
}

int main () {
	freopen("card.in", "r", stdin);
	freopen("card.out", "w", stdout);
	Main::main();
	return 0;
}