模板逆元

Posted 2021-01-10 dukelv

很基础的东西，但是不能马虎，有3种方法，下面一一列举。

一.线性求逆元

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
int n;
ll p,inv[3000005];
void work()
{
    inv[1] = 1;
    duke(i,2,n)
    {
        inv[i] = (p - p / i) * inv[p % i] % p;
    }
    duke(i,1,n)
    {
        printf("%lld
",inv[i]);
    }
}
int main()
{
    read(n);read(p);
    work();
    return 0;
}

二.费马小定理求逆元

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
ll n,p;
ll qpow(ll x,ll y)
{
    ll tot = 1;
    while(y)
    {
        if((y & 1) != 0)
        {
            tot *= x;
        }
        x *= x;
        x %= p;
        tot %= p;
        y >>= 1;
    }
    return tot;
}
int main()
{
    read(n);read(p);
    printf("%lld
",qpow(n,p - 2));
    return 0;
}

三.exgcd求逆元

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(a == 1 && b == 0)
    {
        x = 1;
        y = 0;
        return 1;
    }
    ll t = exgcd(b,a % b,y,x);
    y -= a / b * x;
    return t;
}
int main()
{
    ll n,p,x,y;
    read(n);read(p);
    ll t = exgcd(n,p,x,y);
    printf("%lld
",(x % p + p) % p);
    return 0;
}