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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 |
---|---|---|---|---|---|---|---|---|---|
#537629 | #7512. Almost Prefix Concatenation | OIer_kzc# | WA | 0ms | 3576kb | C++17 | 2.8kb | 2024-08-30 17:00:38 | 2024-08-30 17:00:39 |
Judging History
answer
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
#include <numeric>
#define LOG(FMT...) fprintf(stderr, FMT)
#define eb emplace_back
using namespace std;
typedef long long LL;
constexpr int N = 200005, TC = 30 * N;
int n;
struct Pair {
int a, b, k;
} w[N];
bool cmpA(const Pair &x, const Pair &y) {
return x.a < y.a;
}
bool cmpB(const Pair &x, const Pair &y) {
return x.b < y.b;
}
#define L (tr[z].l)
#define R (tr[z].r)
struct Tree {
int l, r, c; LL v;
} tr[TC];
int rt[N], ids;
void mdf(int &z, int x, int c, int l = 0, int r = n) {
z = ++ids;
tr[z].v = c;
tr[z].c = 1;
if (l == r) {
return;
}
int mid = l + r >> 1;
if (x <= mid) {
mdf(L, x, c, l, mid);
} else {
mdf(R, x, c, mid + 1, r);
}
}
void merge(int &y, int x, int l = 0, int r = n) {
if (!x) {
return;
}
if (!y) {
y = x;
return;
}
tr[y].v += tr[x].v;
tr[y].c += tr[x].c;
if (l == r) {
return;
}
int mid = l + r >> 1;
merge(tr[y].l, tr[x].l, l, mid);
merge(tr[y].r, tr[x].r, mid + 1, r);
}
struct Dat {
int c; LL d;
bool operator < (const Dat &t) const {
return c < t.c || c == t.c && d < t.d;
}
Dat operator + (int tc) const {
return (Dat){c + tc, d};
}
} q[N];
Dat qry(int z, LL v, int l = 0, int r = n) {
if (l == r) {
return (Dat){0, v};
}
int mid = l + r >> 1;
if (tr[L].v <= v) {
return qry(R, v - tr[L].v, mid + 1, r) + tr[L].c;
}
return qry(L, v, l, mid);
}
LL res[N];
void solve(int l, int r, LL vl, LL vr, const vector<int> &v) {
if (l > r) {
return;
}
if (vl == vr) {
for (int k = l; k <= r; ++k) {
res[k] = vr;
}
return;
}
LL md = (vl + vr) >> 1;
int maxc = 0;
for (int i = 0; i < v.size(); ++i) {
int x = v[i];
q[i] = qry(rt[x], md - w[x].b);
maxc = max(maxc, q[i].c);
}
vector<int> nv;
Dat t = (Dat){-1, -1};
for (int i = 0; i < v.size(); ++i) {
if (!(q[i] < t)) {
nv.eb(v[i]);
t = q[i];
}
}
solve(l, maxc, vl, md, nv);
nv.clear();
t = (Dat){-1, -1};
for (int i = v.size() - 1; ~i; --i) {
if (t < q[i]) {
nv.eb(v[i]);
t = q[i];
}
}
reverse(nv.begin(), nv.end());
solve(maxc + 1, r, md + 1, vr, nv);
}
int main() {
scanf("%d", &n);
LL suma = 0; int maxb = 0;
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &w[i].a, &w[i].b);
suma += w[i].a;
maxb = max(maxb, w[i].b);
}
sort(w + 1, w + n + 1, cmpA);
for (int i = 1; i <= n; ++i) {
w[i].k = i - 1;
}
sort(w + 1, w + n + 1, cmpB);
for (int i = 1; i <= n; ++i) {
mdf(rt[i], w[i].k, w[i].a);
merge(rt[i], rt[i - 1]);
}
vector<int> v(n);
iota(v.begin(), v.end(), 1);
solve(1, n, 0, suma + maxb, v);
for (int i = 1; i <= n; ++i) {
printf("%lld\n", res[i]);
}
return 0;
}
详细
Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 3576kb
input:
ababaab aba
output:
result:
wrong answer Answer contains longer sequence [length = 1], but output contains 0 elements