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>
using namespace std;
#define ll long long
#define endl '\n'
#define FF first
#define SS second
#define all(a) a.begin(), a.end()
const int N = 1 << 20;
int n, o, mod;
vector<int> S(2 * N, 1);
int add(int a, int b) {
return (a + b) % mod;
}
int mul(int a, int b) {
a %= mod, b %= mod;
return (ll)a * b % mod;
}
void update(int id, int c) {
id += N;
S[id] += c;
while (id /= 2) {
S[id] = mul(S[id * 2], S[id * 2 + 1]);
}
}
int main() {
ios_base::sync_with_stdio(0);cin.tie(0);
cin >> n >> o >> mod;
vector<array<int, 2>> A(n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < 2; j++) {
cin >> A[i][j];
}
}
sort(all(A));
int r = -1;
while (r + 1 < n && ((2 * A[r + 1][0]) <= A.back()[0])) {
r++;
update(A[r][1], 1);
}
vector<int> Frq(n + 1);
for (int i = 0; i <= n; i++) {
Frq[i] = S[i + N] - 1;
}
vector<int> Last(n + 1, n);
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
int k = A[i][1];
if (Last[k] == n) {
Last[k] = i;
while (r >= 0 && ((2 * A[r][0]) > A[i][0])) {
update(A[r][1], -1);
r--;
}
for (int f = S[k + N]; f > 0; f--) {
if (i == n - 1 || f - 1 == Frq[k]) {
int c = S[k + N];
update(k, -(c - 1));
ans = add(ans, S[1]);
update(k, (c - 1));
}
}
}
}
while (r + 1 < n && ((2 * A[r + 1][0]) <= A.back()[0])) {
r++;
update(A[r][1], 1);
}
for (int i = n - 1; i >= 0; i--) {
int k = A[i][1];
if (Last[k] == i) {
while (r >= 0 && ((2 * A[r][0]) > A[i][0])) {
if (S[A[r][1] + N] != 1)
update(A[r][1], -1);
r--;
}
int b = S[k + N];
update(k, -(S[k + N] - 1));
if (i == n - 1) {
continue;
}
for (int f = b; f > 0; f--) {
if (f - 1 != Frq[k]) {
ans = add(ans, S[1]);
}
}
}
}
cout << ans;
}
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