乘法逆元

ref:http://blog.csdn.net/acdreamers/article/details/8220787

Code(luogu 3811):

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MN 20001005
using namespace std;
inline int in(){
    int x=0;bool f=0; char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
int inv[MN],n,p;
int main()
{
    n=in();p=in();
    inv[1]=1;printf("%d
",inv[1]);
    for (int i=2;i<=n;++i)
    inv[i]=(1ll*(p-(p/i))*inv[p%i])%p,printf("%d
",inv[i]);
    return 0;
}