[Luogu 3811] [TEMPLATE] multiplication inverse element, luogu3811 Multiplication

[Luogu 3811] [TEMPLATE] multiplication inverse element, luogu3811 Multiplication
Background

This is a template question.

Description

Evaluate 1 ~ for given n and p ~ All integers in n represent the multiplication inverse element in the modulo p.

Input/Output Format

Input Format:

N, p in a row

Output Format:

N rows. line I indicates the inverse element of I in the p sense.

Input and Output sample input sample #1:

10 13

Output sample #1:

179108112534

Description

1 ≤ n ≤ 3 × 106, n <p <20000528

Enter pp as the prime number.

Ferma's theorem:

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #define ll long long 6 using namespace std; 7 int n,p; 8 int fast(ll a,ll b){ 9     ll ans=1;10     while(b){11         if(b&1)ans=(ans*a)%p;12         a=(a*a)%p;13         b>>=1;14     }15     return ans;16 }17 int main(){18     scanf("%d%d",&n,&p);19     for(int i=1;i<=n;i++)20         printf("%d\n",fast(i,p-2));21     return 0;22 }

Extended Euclidean:

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 void ex_gcd(int a,int b,int &x,int &y){ 7     if(b==0){x=1,y=0;return;} 8     ex_gcd(b,a%b,x,y); 9     int tmp=x;x=y;y=tmp-(a/b)*y;10 }11 int main(){12     int a,b,x,y;13     scanf("%d%d",&a,&b);14     for(int i=1;i<=a;i++){15         ex_gcd(i,b,x,y);16         printf("%d\n",(x+b)%b);17     }18     return 0;19 }

Linear computing:

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 int ans[3000010],n,p; 7 int main(){ 8     scanf("%d%d",&n,&p); 9     ans[1]=1;puts("1");10     for(int i=2;i<=n;i++){11         ans[i]=(long long)(p-p/i)*ans[p%i]%p;12         printf("%d\n",ans[i]);13     }14     return 0;15 }