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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#257290 | #7749. A Simple MST Problem | ucup-team484# | TL | 826ms | 150772kb | C++17 | 2.2kb | 2023-11-19 02:10:25 | 2023-11-19 02:10:26 |
Judging History
answer
#include <bits/stdc++.h>
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define st first
#define nd second
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
int lp[N], pr[N], cnt = 0, sm[N];
vector<int> divi[N];
void solve() {
int l, r; cin >> l >> r;
ll ans = 0;
vector<vector<pair<int, int>>> arr(r + 1);
for (int i = l; i <= r; i++)
for (int g: divi[i])
arr[g].push_back({sm[i], i});
for (int i = 1; i <= r; i++)
sort(all(arr[i]));
vector<int> par(r + 1);
iota(all(par), 0);
function<int(int)> qry = [&](int x) {
return x == par[x] ? x : par[x] = qry(par[x]);
};
vector<int> ptr(r + 1);
vector<vector<int>> comp(r + 1);
for (int i = l; i <= r; i++)
comp[i].push_back(i);
int cnt = r - l + 1;
while (cnt > 1) {
vector<pair<int, int>> edg;
for (auto& c: comp) {
int opt = mod;
pair<int, int> cand;
for (int x: c) {
for (int g: divi[x]) {
while (ptr[g] < sz(arr[g])) {
if (qry(arr[g][ptr[g]].nd) == qry(x)) {
ptr[g]++;
continue;
}
int val = sm[x] - sm[g] + arr[g][ptr[g]].st;
if (val < opt) {
opt = val;
cand = {x, arr[g][ptr[g]].nd};
}
break;
}
}
}
for (int x: c)
for (int g: divi[x])
ptr[g] = 0;
if (opt != mod)
edg.push_back(cand);
}
for (auto [u, v]: edg) {
if (qry(u) != qry(v))
ans += sm[u] + sm[v] - sm[gcd(u, v)];
u = qry(u);
v = qry(v);
if (u != v) {
if (sz(comp[u]) > sz(comp[v]))
swap(u, v);
comp[v].insert(comp[v].end(), comp[u].begin(), comp[u].end());
comp[u].clear();
cnt--;
par[u] = v;
}
}
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false); cin.tie(0);
for (int i = 2; i < N; i++) {
if (!lp[i])
pr[cnt++] = lp[i] = i, sm[i] = 1;
for (int j = 0; j < cnt && pr[j] * i < N && pr[j] <= lp[i]; j++) {
sm[pr[j] * i] = sm[i] + (pr[j] != lp[i]);
lp[pr[j] * i] = pr[j];
}
}
for (int i = 1; i < N; i++) {
for (int j = i; j < N; j += i)
divi[j].push_back(i);
}
int t; cin >> t; while (t--) solve();
}
Details
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Test #1:
score: 100
Accepted
time: 657ms
memory: 127800kb
input:
5 1 1 4 5 1 4 1 9 19 810
output:
0 2 3 9 1812
result:
ok 5 lines
Test #2:
score: 0
Accepted
time: 818ms
memory: 150772kb
input:
2 27 30 183704 252609
output:
8 223092
result:
ok 2 lines
Test #3:
score: 0
Accepted
time: 826ms
memory: 150712kb
input:
1 183704 252609
output:
223092
result:
ok single line: '223092'
Test #4:
score: -100
Time Limit Exceeded
input:
2 639898 942309 30927 34660