#include<bits/stdc++.h>
using namespace std;
int inv[30000100],n,mod;
int read()
{
char ch=getchar();int k=0;
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') k*=10,k+=ch-'0',ch=getchar();
return k;
}
int main()
{
inv[1]=1;n=read(),mod=read(); cout<<"1"<<endl;
for(int i=2;i<=n;i++) inv[i]=(long long )(mod-mod/i)*inv[mod%i]%mod,inv[i]+=mod,inv[i]%=mod,printf("%d\n",inv[i]);
}
乘法逆元
最新推荐文章于 2022-07-15 19:10:04 发布