#if 0 おそらく AC 解 #endif // includes {{{ #include #include #include #include #include #include #include #include #include #include #include #include #include #include // #include // #include // #include // #include // }}} using namespace std; using ll = long long; // .add(i, v) : bit[i] += v // .get(i) : bit[i] // .sum(i) : bit[0] + ... + bit[i] // .range(l, r) : bit[l] + ... + bit[r] // .lower_bound(T v) : min i that satisfies .sum(i) >= v // - use only when bit[i] >= 0 for all i > 0 /// --- Binary Indexed Tree {{{ /// #include #include template < class T = long long > struct BinaryIndexedTree { using size_type = std::size_t; size_type n, m; T identity; std::vector< T > data; BinaryIndexedTree() : n(0) {} BinaryIndexedTree(int n, T identity = T()) : n(n), identity(identity), data(n, identity) { m = 1; while(m < n) m <<= 1; } void add(size_type i, T x) { assert(i < n); i++; while(i <= n) { data[i - 1] = data[i - 1] + x; i += i & -i; } } T sum(int i) { if(i < 0) return identity; if(i >= n) i = n - 1; i++; T s = identity; while(i > 0) { s = s + data[i - 1]; i -= i & -i; } return s; } T get(int i) { return sum(i) - sum(i - 1); } T range(int a, int b) { return sum(b) - sum(a - 1); } size_type lower_bound(T w) { size_type i = 0; for(size_type k = m; k > 0; k >>= 1) { if(i + k <= n && data[i + k - 1] < w) w -= data[(i += k) - 1]; } return i; } }; /// }}}--- /// template < class T = long long > using BIT = BinaryIndexedTree< T >; int main() { std::ios::sync_with_stdio(false), std::cin.tie(0); int n, q; cin >> n >> q; assert(1 <= n and n <= int(1e5)); assert(1 <= q and q <= int(1e5)); vector a(n); vector> v; for(int i = 0; i < n; i++) { cin >> a[i], v.emplace_back(a[i], i); assert(int(-1e9) <= a[i] and a[i] <= int(1e9)); } sort(begin(v), end(v)); vector l(q), r(q); for(int i = 0; i < q; i++) { cin >> l[i] >> r[i]; assert(1 <= l[i] and l[i] <= r[i] and r[i] <= n); l[i]--, r[i]--; } vector ok(q, n-1); vector ng(q, -1); if(n > 1) { vector> mid(n); mid[n/2].reserve(n); for(int i = 0; i < q; i++) mid[n/2].push_back(i); // パラサーチ O(Q log N) * O(log N) int rest = q; while(rest) { BIT<> bit(n); for(int i = 0; i < n; i++) { int id = v[i].second; bit.add(id, 1); for(int j : mid[i]) { if(bit.range(l[j], r[j]) >= (r[j] - l[j] + 2) / 2) ok[j] = i; else ng[j] = i; if(abs(ok[j] - ng[j]) > 1) { mid[(ok[j] + ng[j])/2].push_back(j); } else { rest--; } } mid[i].clear(); } } } vector> qs(n); for(int i = 0; i < q; i++) qs[ok[i]].push_back(i); // dump(ok); BIT<> d1(n), d2(n); vector ans(q); for(int i = 0; i < n; i++) d2.add(i, a[i]); for(int i = 0; i < n; i++) { int id = v[i].second; d2.add(id, -a[id]); d1.add(id, a[id]); for(auto j : qs[i]) { int len = r[j] - l[j] + 1; ans[j] = - d1.range(l[j], r[j]) + d2.range(l[j], r[j]); if(len % 2 == 1) ans[j] += a[id]; // dump(j, len, a[id]); } } for(int i = 0; i < q; i++) cout << ans[i] << "\n"; return 0; }