이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
// author: aykhn
using namespace std;
typedef long long ll;
#define TC int t; cin >> t; while (t--) _();
#define OPT ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(v) v.begin(), v.end()
#define pii pair<int, int>
#define mpr make_pair
#define eb emplace_back
#define pdd pair<double, double>
#define pb push_back
#define ts to_string
#define fi first
#define se second
#define int ll
#define ins insert
#define inf 0x3F3F3F3F
#define infll 0x3F3F3F3F3F3F3F3FLL
#define bpc __builtin_popcount
struct custom_hash
{
static uint64_t splitmix64(uint64_t x)
{
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
int n;
int a[2][100001];
struct SegTree
{
int sz;
vector<int> st;
void build(int l, int r, int x, int id)
{
if (l + 1 == r)
{
if (l < n) st[x] = a[id][l];
return;
}
int mid = (l + r) >> 1;
build(l, mid, 2*x + 1, id);
build(mid, r, 2*x + 2, id);
st[x] = st[2*x + 1] + st[2*x + 2];
}
void build(int id)
{
sz = 1;
while (sz < n) sz <<= 1;
st.assign(sz << 1, 0);
build(0, sz, 0, id);
}
void make(int ind, int val, int l, int r, int x)
{
if (l + 1 == r)
{
st[x] = val;
return;
}
int mid = (l + r) >> 1;
if (ind < mid) make(ind, val, l, mid, 2*x + 1);
else make(ind, val, mid, r, 2*x + 2);
st[x] = st[2*x + 1] + st[2*x + 2];
}
void make(int ind, int val)
{
make(ind - 1, val, 0, sz, 0);
}
int get(int lx, int rx, int l, int r, int x)
{
if (l >= rx || r <= lx) return 0;
if (l >= lx && r <= rx) return st[x];
int mid = (l + r) >> 1;
return get(lx, rx, l, mid, 2*x + 1) + get(lx, rx, mid, r, 2*x + 2);
}
int get(int l, int r)
{
if (l > r) return 0;
return get(l - 1, r, 0, sz, 0);
}
};
SegTree st[2];
int solve1(int l, int r, int t)
{
if (l > r) return 0;
int r1 = l + t;
int len = min(r + 1, r1) - l;
int s1 = st[1].get(l, l + len - 1) - st[0].get(l, l + len - 1) * (l - 1);
int s2 = st[0].get(r1, r) * t;
return s1 + s2;
}
int solve2(int l, int r, int t)
{
if (l > r) return 0;
int l1 = r - t;
int len = r - max(l - 1, l1);
int s1 = st[0].get(r - len + 1, r) * (r + 1) - st[1].get(r - len + 1, r);
int s2 = st[0].get(l, l1) * t;
return s1 + s2;
}
signed main()
{
OPT;
int k;
cin >> n >> k;
int b[k + 1];
for (int i = 0; i < n; i++)
{
cin >> a[0][i];
a[1][i] = a[0][i] * (i + 1);
}
st[0].build(0);
st[1].build(1);
int q;
cin >> q;
while (q--)
{
int type;
cin >> type;
if (type == 1)
{
for (int j = 1; j <= k; j++)
{
cin >> b[j];
b[j]--;
if (j > 1)
{
swap(a[0][b[j - 1]], a[0][b[j]]);
}
}
for (int j = 1; j <= k; j++)
{
a[1][b[j]] = a[0][b[j]] * (b[j] + 1);
st[0].make(b[j] + 1, a[0][b[j]]);
st[1].make(b[j] + 1, a[1][b[j]]);
}
}
else
{
int l, r, m;
cin >> l >> r >> m;
int sz = r - l + 1;
int t = sz - m + 1;
int res = solve1(l, l + (sz + 1)/2 - 1, min(t, m)) + solve2(l + (sz + 1)/2, r, min(t, m));
cout << res << endl;
}
}
}
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