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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 |
---|---|---|---|---|---|---|---|---|---|
#876813 | #5. 在线 O(1) 逆元 | atgc | 80 | 5340ms | 4608kb | C++14 | 529b | 2025-01-31 13:27:05 | 2025-01-31 13:27:06 |
Judging History
answer
// #include<bits/stdc++.h>
// using namespace std;
#include"inv.h"
const int maxn = 2e5+10,mod=998244353;
int inv_a[maxn];
void init(int){
inv_a[1]=1;for(int i=2;i<maxn;++i)inv_a[i]=(long)(mod-mod/i)*inv_a[mod%i]%mod;
}
int inv(int x){
long scal=1;
while(x>=maxn)(scal*=-mod/x)%=mod,x=mod%x;
return(scal+mod)*inv_a[x]%mod;
}
// signed main() {
// ios::sync_with_stdio(0),cin.tie(0);
// init(0);
// while(1){
// int z=rand()%mod+1;
// assert((long)z*inv(z)%mod == 1);
// cerr<<z;
// }
// }
详细
Pretests
Final Tests
Test #1:
score: 10
Accepted
time: 6ms
memory: 4480kb
Test #2:
score: 20
Accepted
time: 675ms
memory: 4608kb
Test #3:
score: 30
Accepted
time: 3355ms
memory: 4352kb
Test #4:
score: 20
Accepted
time: 5340ms
memory: 4608kb
Test #5:
score: 0
Time Limit Exceeded