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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#641044 | #5. 在线 O(1) 逆元 | SkyMaths | 80 | 5184ms | 42944kb | C++14 | 849b | 2024-10-14 18:03:14 | 2024-11-05 22:06:14 |
Judging History
answer
#pragma GCC optimize("-Ofast", "-finline", "-funroll-loops", "-fno-stack-protector")
#include<bits/stdc++.h>
// #include "inv.h"
#define rep(i,l,r) for (int i(l), i##end(r); i <= i##end; ++i)
#define per(i,r,l) for (int i(r), i##end(l); i >= i##end; --i)
using namespace std;
const int N = 1e7 + 9, Mod = 998244353;
int n = N - 9, MMI[N];
void init(int p) {
MMI[1] = 1;
rep (i, 2, n) MMI[i] = Mod - 1ll * (Mod / i) * MMI[Mod % i] % Mod;
}
int inv(int x) {
if (x <= n) return MMI[x];
int q = Mod / x, r = Mod % x;
if (r < x / 2) return Mod - 1ll * q * inv(r) % Mod;
return 1ll * (q + 1) * inv(x - r) % Mod;
}
// const int inf = 1e9;
// mt19937 mtrnd(time(0));
// signed main() {
// init(Mod);
// rep (t, 1, 10) {
// int x = mtrnd() % inf + 1;
// assert(1ll * x * inv(x) % Mod == 1);
// }
// return 0;
// }
Details
Pretests
Final Tests
Test #1:
score: 10
Accepted
time: 65ms
memory: 42752kb
Test #2:
score: 20
Accepted
time: 703ms
memory: 42944kb
Test #3:
score: 30
Accepted
time: 3264ms
memory: 42784kb
Test #4:
score: 20
Accepted
time: 5184ms
memory: 42804kb
Test #5:
score: 0
Time Limit Exceeded