[P3811][模板]乘法逆元

原题链接

我只是想打个小费马的板子

结果这个题用小费马过不了[气哭]
又跑去看了半天递推式

总之就是

inv[i]=(p−pi)×inv[p%i]%p

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<ctime>#include<cstdlib>#include<climits>#include<queue>#include<vector>#define LL long long#define MOD 1000000007using namespace std;LL n,p,arc[3000005],i;int main(){    scanf("%lld%lld",&n,&p);    printf("1\n");    arc[1]=1;    for(i=2;i<=n;i++)    {        arc[i]=(p-p/i)*arc[p%i]%p;        printf("%lld\n",arc[i]);    }    return 0;}