【模板】乘法逆元

扩欧求逆元：

#include<iostream>
#include<cstdio>
using namespace std;

int n, b;

int ex_gcd(int a, int b, int &x, int &y) {
    if(b == 0) {
        x = 1, y = 0;
        return a;
    }
    int g = ex_gcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
}

int main() {
    cin >> n >> b;
    for(int a = 1; a <= n; a++) {
        int x, y;
        int g = ex_gcd(a, b, x, y);
        int t = b / g;
        x = (x % t + t) % t;
        printf("%d
", x);
    }
}

线性递推求逆元：

#include<iostream>
#include<cstdio>
using namespace std;

const int MAXN = 3000000 + 5;

int n, p;
int iel[MAXN];

int main() {
    iel[1] = 1;
    scanf("%d%d", &n, &p);
    for(int i = 2; i <= n; i++) {
        //被注释掉的写法和程序中的写法是等价的
        //iel[i] = 1LL * (p - p / i) * iel[p % i] % p;
        iel[i] = - 1LL * (p / i) * iel[p % i] % p;
        iel[i] = ((iel[i]) % p + p) % p;
    }
    for(int i = 1; i <= n; i++)
        printf("%d
", iel[i]);
}

线性求逆元

要下晚自习了..证明明天补...