洛谷P3811乘法逆元

传送门

线性递推

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define re register
using namespace std;
const int maxn = 3000005;

int n;
long long mod;
int p[maxn];

int main(){
	scanf("%d%lld",&n,&mod);
	printf("%d
",p[1] = 1);
	for(re int i = 2 ; i <= n ; ++i) {
		p[i] = -mod / i * p[mod % i] % mod;
		if(p[i] < 0) p[i] += mod;
		printf("%d
",p[i]);
	}
	return 0;
}

快速幂

#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

ll n,p;

ll ksm (ll a,ll b=p-2){
    ll ans=1;
    while(b>0){
        if(b&1){
            ans=ans*a%p;
        }
        a=a*a%p;
        b>>=1; 
    }
    return ans;
} 

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>p;
    for(int i=1;i<=n;i++) {
        cout<<ksm(i)<<endl;
    }
    return 0;
}

拓展欧几里得

#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

ll n,p;
ll x,y;

void exgcd(ll a,ll b,ll &x,ll &y){
    if(b==0){
        x=1,y=0;
        return;
    }
    exgcd(b,a%b,y,x);
    y-=a/b*x;
} 

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>p;
    for(int i=1;i<=n;i++){
        exgcd(i,p,x,y);
        cout<<((x%p)+p)%p<<endl;    
    }
    return 0;
}