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ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#106139 | #5. 在线 O(1) 逆元 | zhoukangyang | Compile Error | / | / | C++17 | 1.2kb | 2023-05-16 18:25:13 | 2024-11-05 21:49:34 |
Judging History
你现在查看的是最新测评结果
- [2024-11-05 21:49:34]
- 管理员手动重测本题所有提交记录
- 测评结果:Compile Error
- 用时:0ms
- 内存:0kb
- [2023-08-10 23:21:45]
- System Update: QOJ starts to keep a history of the judgings of all the submissions.
- [2023-05-16 18:25:16]
- 评测
- 测评结果:Compile Error
- 用时:0ms
- 内存:0kb
- [2023-05-16 18:25:13]
- 提交
answer
#include<bits/stdc++.h>
#include "inv.h"
#define L(i, j, k) for(int i = (j); i <= (k); ++i)
#define R(i, j, k) for(int i = (j); i >= (k); --i)
#define ll long long
#define vi vector < int >
#define sz(a) ((int) (a).size())
#define ll long long
#define ull unsigned long long
#define me(a, x) memset(a, x, sizeof(a))
#define eb emplace_back
using namespace std;
const int B = 1024, buf = 3, mod = 998244353, N = 6e6 + 7;
int K, ml[N], iv[N], dis[N];
int Inv(int x) {
return x < N ? iv[x] : (ll) Inv(mod % x) * (mod - mod / x) % mod;
}
int inv(int w) {
int d = w >> 10, M = ml[d];
return M ? Inv(w) : (ll) iv[w * M - dis[d]] * M % mod;
}
void init(int rp) {
K = mod / B;
L(fb, 1, B * buf) {
int cur = 0, Add = fb * B;
for(int p = 0; p <= K; ) {
if(cur <= K * buf) {
ml[p] = fb;
cur += Add, ++p;
} else {
int A = (mod - cur + Add - 1) / Add;
cur += A * Add, p += A;
}
if(cur >= mod) cur -= mod;
}
}
// int count = 0;
L(i, 1, K) if(ml[i]) dis[i] = (ll) ml[i] * i * B / mod * mod, ++count;
// cout << 1. * count / K << endl;
iv[1] = 1;
L(i, 2, N - 1) iv[i] = (ll) iv[mod % i] * (mod - mod / i) % mod;
}
Details
implementer.cpp: In function ‘int main()’: implementer.cpp:22:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 22 | scanf("%d", &n); | ~~~~~^~~~~~~~~~ answer.code: In function ‘void init(int)’: answer.code:38:73: error: no pre-increment operator for type 38 | L(i, 1, K) if(ml[i]) dis[i] = (ll) ml[i] * i * B / mod * mod, ++count; | ^~~~~