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QOJ
| 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 |
|---|---|---|---|---|---|---|---|---|---|
| #534330 | #5. 在线 O(1) 逆元 | surgutti | 0 | 0ms | 0kb | C++14 | 363b | 2024-08-27 02:43:15 | 2024-11-05 22:03:22 |
Judging History
answer
#include "inv.h"
constexpr int N = 200000000 + 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 - (long long) (mod / i) * inverse[mod % i] % mod;
}
int inv(int x) {
if (x < N)
return inverse[x];
return mod - (long long) (mod / x) * inv(mod % x) % mod;
}
詳細信息
Pretests
Final Tests
Test #1:
score: 0
Memory Limit Exceeded
Test #2:
score: 0
Memory Limit Exceeded
Test #3:
score: 0
Memory Limit Exceeded
Test #4:
score: 0
Memory Limit Exceeded
Test #5:
score: 0
Time Limit Exceeded