QOJ.ac
QOJ
| ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
|---|---|---|---|---|---|---|---|---|---|
| #534328 | #5. 在线 O(1) 逆元 | surgutti | 0 | 985ms | 394500kb | C++14 | 351b | 2024-08-27 02:41:46 | 2024-11-05 22:03:20 |
Judging History
answer
#include "inv.h"
constexpr int N = 100000000 + 1;
constexpr int mod = 998244353;
int inverse[N];
void init(int p) {
inverse[1] = 1;
for (int i = 2; i < N; i++)
inverse[i] = (mod - mod / i * inverse[mod % i] % mod) % mod;
}
int inv(int x) {
if (x < N)
return inverse[x];
return (mod - mod / x * inv(mod % x) % mod) % mod;
}
Details
Pretests
Final Tests
Test #1:
score: 0
Wrong Answer
time: 945ms
memory: 394368kb
Test #2:
score: 0
Wrong Answer
time: 985ms
memory: 394500kb
Test #3:
score: 0
Wrong Answer
time: 965ms
memory: 394372kb
Test #4:
score: 0
Wrong Answer
time: 977ms
memory: 394412kb
Test #5:
score: 0
Wrong Answer
time: 931ms
memory: 394252kb