QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#643732 | #7789. Outro: True Love Waits | Mcggvc | RE | 0ms | 0kb | C++17 | 4.1kb | 2024-10-15 23:33:03 | 2024-10-15 23:33:05 |
answer
#include <bits/stdc++.h>
using namespace std;
// typedef long long lld;
#define int long long
const int N = 100005;
struct SGT {
int d[N << 2], b[N << 2], v[N << 2], n;
void build(int s, int t, int p, int *a) {
// 对 [s,t] 区间建立线段树,当前根的编号为 p
if (s == t) {
d[p] = a[s];
return;
}
int m = s + ((t - s) >> 1);
// 移位运算符的优先级小于加减法,所以加上括号
// 如果写成 (s + t) >> 1 可能会超出 int 范围
build(s, m, p * 2, a), build(m + 1, t, p * 2 + 1, a);
// 递归对左右区间建树
d[p] = d[p * 2] + d[(p * 2) + 1];
}
void update(int l, int r, int c, int s, int t, int p) {
if (l <= s && t <= r) {
d[p] = (t - s + 1) * c, b[p] = c, v[p] = 1;
return;
}
int m = s + ((t - s) >> 1);
// 额外数组储存是否修改值
if (v[p]) {
d[p * 2] = b[p] * (m - s + 1), d[p * 2 + 1] = b[p] * (t - m);
b[p * 2] = b[p * 2 + 1] = b[p];
v[p * 2] = v[p * 2 + 1] = 1;
v[p] = 0;
}
if (l <= m) update(l, r, c, s, m, p * 2);
if (r > m) update(l, r, c, m + 1, t, p * 2 + 1);
d[p] = d[p * 2] + d[p * 2 + 1];
}
int getsum(int l, int r, int s, int t, int p) {
if (l <= s && t <= r) return d[p];
int m = s + ((t - s) >> 1);
if (v[p]) {
d[p * 2] = b[p] * (m - s + 1), d[p * 2 + 1] = b[p] * (t - m);
b[p * 2] = b[p * 2 + 1] = b[p];
v[p * 2] = v[p * 2 + 1] = 1;
v[p] = 0;
}
int sum = 0;
if (l <= m) sum = getsum(l, r, s, m, p * 2);
if (r > m) sum += getsum(l, r, m + 1, t, p * 2 + 1);
return sum;
}
int getpos(int x) {
if(x == 1) {
return getsum(1, 1, 1, n, 1);
} else {
return getsum(1, x, 1, n, 1) - getsum(1, x - 1, 1, n, 1);
}
}
int sum() {
return getsum(1, n, 1, n, 1);
}
void clear() {
for(int i = 0; i < n * 4 + 1; i++) {
d[i] = b[i] = v[i] = 0;
}
n = 0;
}
};
int find(int x, int flg, SGT &T) {
// if(x == T.n) return x;
int l = x + 1, r = T.n;
while(l < r) {
int mid = (l + r + 1) >> 1;
if(T.getpos(mid) > flg) {
r = mid - 1;
} else {
l = mid;
}
}
// if(l == T.n && T.getpos(T.n) <= flg) {
// return T.n;
// }
if(T.getpos(r) > flg) return r - 1;
return r;
}
SGT F, G;
int f[N], g[N];
int a[N], suma = 0, maxa = 0;
void solve() {
int n; cin >> n;
suma = 0, maxa = 0;
for(int i = 1; i <= n; i++) {
cin >> a[i];
suma += a[i]; maxa = max(maxa, a[i]);
}
f[0] = 0;
for(int i = 1; i <= n; i++) {
f[i] = max(f[i - 1], a[i]);
}
g[n + 1] = 0;
for(int j = n; j >= 1; j--) {
g[j] = max(g[j + 1], a[j]);
}
reverse(g + 1, g + 1 + n);
F.n = n; G.n = n;
F.build(1, n, 1, f); G.build(1, n, 1, g);
int q; cin >> q;
while(q--) {
int x, v; cin >> x >> v;
suma += v;
maxa = max(maxa, a[x] + v);
a[x] += v;
int nxt = 0;
if(a[x] > F.getpos(x)) {
nxt = find(x, a[x], F);
F.update(x, nxt, a[x], 1, n, 1);
}
if(a[x] > G.getpos(n - x + 1)) {
nxt = find(n - x + 1, a[x], G);
G.update(n - x + 1, nxt, a[x], 1, n, 1);
}
// cout << "FSUM : " << F.sum() << " GSUM : " << G.sum() << "\n";
cout << F.sum() + G.sum() - n * maxa - suma << "\n";
// for(int i = 1; i <= n; i++) {
// cout << F.getpos(i) << " ";
// }
// cout << "\n";
}
F.clear(); G.clear();
}
signed main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int T; cin >> T;
while(T--) solve();
return 0;
}
詳細信息
Test #1:
score: 0
Runtime Error
input:
4 1 10 1 1 10 2 100 0 2 11 11 3