C. Divan and bitwise operations

// Problem: C. Divan and bitwise operations
// Contest: Codeforces - Codeforces Round #757 (Div. 2)
// URL: https://codeforces.com/contest/1614/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-02-16 00:46:49
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}int n, m;
const int mod = 1e9 + 7;
ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a %mod;
		a= a * a%mod;
		b >>= 1;
	}
	return ans;
}
void solve() {
	cin >> n >> m;
	ll x= 0;
	for (int i = 1; i <= m; i++) {
		int l, r;
		cin >> l >> r;
		int t;
		cin >> t;
		x |= t;
	}
	cout << x * qmi(2, n - 1) %mod << endl;
}
int main () {
    int t;
    cin >> t;
    while (t --) solve();
    return 0;
}

I wanted to pass this question at first l,r Construct an array , Then find the subsequence result . But constructing an array is too complicated . Because we finally find the XOR sum of all subsequences , Or no additional contribution 1, So don't consider the specific number . If we've got a bunch of numbers , For someone , We ask for his total contribution to all sequences , We divide numbers into A,B Two sets ,A Indicates this bit 0,B Indicates that this digit is 1, Result If this one is 0, Then we choose 0,1,2,.... Are all the same , Because XOR 0 For the original number face 1, We can only choose an odd number of times , Because even times is 0, Only an odd number of times can you contribute . Then the result is 2^(m - 1) The total contribution is 2^(A) * 2^(B - 1) = 2^(n -1) Then by recording the weight of each bit . Of x. The final result is x*(2^(n - 1)); There is no need to consider specific arrays .