## Number of single O (logmod) inversing

\$View\$ \$Code\$
```

const int mod=998244353;
inline long long qpow(long long a,long long b)
{
long long ans=1;
while(b)
{
if(b&1)
ans=ans*a%mod;
a=a*a%mod;
b/=2;
}
return ans%mod;
}
long long calcinv(long long x)
{
return qpow(x,mod-2);
}

```

## O (n) to find the inverse element 1 ~ n (linear inverse element)

\(View\) \(Code\)

``````#include<bits/stdc++.h>
using namespace std;
{
int ret=0,f=1;
char ch=getchar();
while('9'<ch||ch<'0')
{
if(ch=='-')
f=-1;
ch=getchar();
}
while('0'<=ch&&ch<='9')
{
ret=(ret<<1)+(ret<<3)+ch-'0';
ch=getchar();
}
return ret*f;
}
int n,p;
long long inv[3000005];
int main()
{
inv[1]=1;
for(register int i=2;i<=n;i++)
inv[i]=1ll*(p-p/i)*inv[p%i]%p;
for(register int i=1;i<=n;i++)
printf("%lld\n",inv[i]);
return 0;
}``````

## O (n) 1 to find the inverse element n factorial (factorial inverse element)

\$View\$ \$Code\$
```

const int mod=998244353;
int maxn;
long long ans,fac[100005],inv[100005];
inline void pre()
{
fac[1]=1;
inv[0]=1;
for(register int i=2;i<=maxn;i++)
{
fac[i]=fac[i-1]*i%mod;
}
inv[maxn]=qpow(fac[maxn],mod-2,mod);
for(register int i=maxn;i;i--)
{
inv[i-1]=inv[i]*i%mod;
}
}

```

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Origin www.cnblogs.com/Peter0701/p/11869760.html
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