$Theta(n)$求$1...n$的乘法逆元

Posted 2021-03-14 chz-hc

设 [ p = m * d + r]

也就是(p)为被除数，(d)为除数，(m)为商，(r)为余数

[ m = [frac{p}{d}]]

[ r = p mod d]

[ m * d + r equiv 0mod p]

两边同乘(d^{-1}r^{-1})得

[m * r^{-1} + d^{-1} equiv 0mod p]

移项得

[ d^{-1} equiv -m * r^{-1}mod p]

因为 [m = [frac{p}{d}]]

所以[d^{-1} equiv -[frac{p}{d}] * r^{-1}]

因为 [ r = pmod d]

[ d^{-1} equiv -[frac{p}{d}]*[pmod d]^{-1} mod p]

我们可以记个数组(ans[])

初始把(ans[1] = 1)

然后根据前面的(ans[pmod d])推出后面的(ans[d])

Code

#include <bits/stdc++.h>

#define ll long long
const int MaxN = 3000000 + 10;

using namespace std;

inline int read() {
    int cnt = 0, opt = 1;
    char ch = getchar();

    for (; ! isalnum(ch); ch = getchar())
        if (ch == '-')  opt = 0;
    for (; isalnum(ch); ch = getchar())
        cnt = cnt * 10 + ch - 48;

    return opt ? cnt : -cnt;
}

ll n, p;
ll ans[MaxN];

int main() {
    n = read(), p = read();

    ans[1] = 1;

    printf("%lld
", ans[1]);

    for (int i = 2; i <= n; ++ i) {
        ans[i] = (p - (p / i)) * ans[p % i] % p;
        printf("%lld
", ans[i]);
    }
    return 0;
}