이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,lzcnt,mmx,abm,avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define tp3 tuple<int, int, int>
const int N = 31, mod = 1e9+7;
int n, q, a[N], b[N], f[N];
tp3 queries[N];
/*
dp[i][j] de cuantas formas puedo ir al array i con las j primeras queries
*/
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
cin >> q;
for (int i = 0; i < q; i++) {
int l, r, x;
cin >> l >> r >> x;
l--; r--;
queries[i] = make_tuple(l, r, x);
}
int ans = 0;
for (int mask = 0; mask < (1 << q); mask++) {
for (int i = 0; i < n; i++) {
f[i] = 0;
b[i] = a[i];
}
for (int i = 0; i < q; i++) {
if ((mask >> i) & 1) {
int l = get<0>(queries[i]);
int r = get<1>(queries[i]);
int x = get<2>(queries[i]);
f[l] ^= x;
f[r+1] ^= x;
}
}
for (int i = 0; i < n; i++) {
if (i) f[i] ^= f[i-1];
b[i] ^= f[i];
}
for (int i = 0; i < n; i++) {
int sum = 0;
for (int j = i; j < n; j++) {
sum += b[j];
ans += sum*sum;
ans %= mod;
}
}
}
cout << ans << "\n";
}
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