제출 #530911

#제출 시각아이디문제언어결과실행 시간메모리
530911abc864197532Intergalactic ship (IZhO19_xorsum)C++17
47 / 100
2058 ms2504 KiB
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define mp make_pair
#define eb emplace_back
#define pb push_back
#define pii pair<int,int>
#define X first
#define Y second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
void abc() {cout << endl;}
template <typename T, typename ...U> void abc(T i, U ...j) {
	cout << i << ' ', abc(j...);
}
template <typename T> void printv(T l, T r) {
	for (; l != r; ++l)
		cout << *l << " \n"[l + 1 == r];
}
#ifdef Doludu
#define test(x...) abc("[" + string(#x) + "]", x);
#define owo freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#else
#define test(x...) void(0)
#define owo ios::sync_with_stdio(false)
#endif
const int N = 1005, mod = 1e9 + 7, inv = (mod + 1) / 2, C = 128;

void trans (vector <lli> &dp, int x) {
	vector <lli> pre = dp;
	for (int i = 0; i < C; ++i) {
		dp[i ^ x] += pre[i];
		if (dp[i ^ x] >= mod)
			dp[i ^ x] -= mod;
	}
}

int main () {
	owo;
	int n, q;
	cin >> n;
	vector <int> a(n);
	for (int i = 0; i < n; ++i)
		cin >> a[i];
	cin >> q;
	vector <array <int, 3>> queries(q);
	for (int i = 0; i < q; ++i) {
		cin >> queries[i][0] >> queries[i][1] >> queries[i][2];
		--queries[i][0];
	}
	lli ans = 0;
	for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) {
		lli pre = 0;
		vector <lli> dp1(C, 0), dp2(C, 0), dp3(C, 0);
		dp1[a[i]] = 1, dp2[a[j]] = 1, dp3[0] = 1;
		lli pw = 1;
		for (auto [l, r, x] : queries) {
			bool inter1 = l <= i && i < r, inter2 = l <= j && j < r;
			if (inter1 && inter2) {
				trans(dp3, x);
			} else if (inter1) {
				trans(dp1, x);
			} else if (inter2) {
				trans(dp2, x);
			} else {
				pw = pw * 2 % mod;
			}
		}
		for (int i = 0; i < C; ++i) {
			lli A = 0, B = 0;
			for (int j = 0; j < C; ++j) {
				(A += dp1[i ^ j] * j) %= mod;
				(B += dp2[i ^ j] * j) %= mod;
			}
			(pre += A * B % mod * dp3[i]) %= mod;
		}
		int ni = i, nj = j;
		if (ni > nj)
			swap(ni, nj);
		(ans += pre * pw % mod * (ni + 1) % mod * (n - nj)) %= mod;
	}
	cout << ans << '\n';
}
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