LuoguP5390 [Cnoi2019]数学作业(数论)

转进制，然后发现贡献只有(1_{(2)})，取奇数个的子集方案是(2^{n-1})

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e_Line printf("
---------------
")
#define D_e(x) cout << (#x) << " : " << x << "
"
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define TIME() fprintf(stderr, "
TIME : %.3lfms
", clock() * 1000.0 / CLOCKS_PER_SEC)
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#define FileSave() ;
#define TIME() ;
#endif
struct ios {
	template<typename ATP> inline ios& operator >> (ATP &x) {
		x = 0; int f = 1; char c;
		for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
		while(c >= '0' && c <='9') x = x * 10 + (c ^ '0'), c = getchar();
		x *= f;
		return *this;
	}
}io;
using namespace std;
template<typename ATP> inline ATP Max(ATP a, ATP b) {
	return a > b ? a : b;
}
template<typename ATP> inline ATP Min(ATP a, ATP b) {
	return a < b ? a : b;
}
template<typename ATP> inline ATP Abs(ATP a) {
	return a < 0 ? -a : a;
}

const int mod = 998244353;

long long mul(long long a, long long b) {
    long long l = a * (b >> 25ll) % mod * (1ll << 25) % mod;
    long long r = a * (b & ((1ll << 25) - 1)) % mod;
    return (l + r) % mod;
}
inline long long Pow(int a, long long b) {
	long long s = 1;
	while(b){
		if(b & 1) s = mul(s, a);//s * a % mod;
		a = mul(a, a), b >>= 1;//a = a * a % mod, b >>= 1;
	}
	return s;
}
int main() {
	int Tasks;
	io >> Tasks;
	while(Tasks--){
		int n;
		io >> n;
		long long ans = 0;
		R(i,1,n){
			int x;
			io >> x;
			ans |= x;
		}
		printf("%lld
", ans * Pow(2, n - 1) % mod);
	
	}
	return 0;
}