c++ 求逆元代码

第一种 利用前缀求n个数的逆元

   

#include<bits/stdc++.h>
#include<unordered_map>
#include<math.h>
#include<forward_list>
#include<unordered_set>
using namespace std;
#pragma warning(disable:4996)
const double pi = acos(-1);
#define pdd pair<double,double>
#define re register
#define ll long long
#define endl '\n'
#define pii pair<ll,ll>
#define IOS std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ull unsigned ll
#define eb emplace_back
#define pb push_back
#define rad read
#define mp make_pair
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
#define fpn(x) fp(x),putchar('\n');
ll mod = 80112002;
inline ll  gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
inline ll exgcd(ll a, ll b, ll& x, ll& y) {
	//ax+by=1; 返回值为a,b最大公约数
	if (!b) {
		x = 1, y = 0;
		return a;
	}
	ll g= exgcd(b, a % b, y, x);
	y -= (a / b) * x;
	return g;
}
inline ll ksc(ll a, ll b, ll mod) {

	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b, ll mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)
			ans = ksc(ans, a, mod);
		a = ksc(a, a, mod);
		b >>= 1;
	}
	return ans;

}
inline ll read() {
	ll s = 0; bool w = true;
	char c = getchar();
	for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
	for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
	return w ? s : -s;
}
inline void fp(ll n) {
	if (n < 0)  putchar('-'), n = -n;
	if (n > 9) fp(n / 10);
	putchar(n % 10 + 48);
}
const int N = 3e6 + 10;
ll inv[N], s[N],a[N];
void solve() {
	ll n = read();
	for (int i = 1; i <= n; ++i)a[i] = read();//求a1-an的逆元
	ll p = read();
	inv[1] = s[1] = 1;
	for (int i = 2; i <= n; ++i)
		s[i] = (s[i - 1] % p * a[i] % p) % p;//改为 s[i - 1] * i 的话，下面也要改  
	inv[n] = qpow(s[n], p - 2, p);
	for (int i = n - 1; i >= 1; --i)
		inv[i] = (inv[i + 1] % p * a[i+1] % p) % p;// inv[i + 1]  * (i + 1) 
	for (int i = 2; i <= n; ++i)
		inv[i] = (inv[i] % p * s[i - 1] % p) % p;
	for (int i = 1; i <= n; ++i)fpn(inv[i]);
}
signed main()
{
	
	ll T = 1;
	//   T = read();
	while (T--) {

		solve();
	}
	return 0;
}
/**
*　　┏┓  WQJ ┏┓+ +
*　┏┛┻━━━┛┻┓ +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　 　　　    ┣┓>>>
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/

第二种 线性求逆元（1-n）洛谷模板题

#include<bits/stdc++.h>
#include<unordered_map>
#include<math.h>
#include<forward_list>
#include<unordered_set>
using namespace std;
#pragma warning(disable:4996)
const double pi = acos(-1);
#define pdd pair<double,double>
#define re register
#define ll long long
#define endl '\n'
#define pii pair<ll,ll>
#define IOS std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ull unsigned ll
#define eb emplace_back
#define pb push_back
#define rad read
#define mp make_pair
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
#define fpn(x) fp(x),putchar('\n');
ll mod = 80112002;
inline ll  gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
inline ll exgcd(ll a, ll b, ll& x, ll& y) {
	//ax+by=1; 返回值为a,b最大公约数
	if (!b) {
		x = 1, y = 0;
		return a;
	}
	ll g= exgcd(b, a % b, y, x);
	y -= (a / b) * x;
	return g;
}
inline ll ksc(ll a, ll b, ll mod) {

	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b, ll mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)
			ans = ksc(ans, a, mod);
		a = ksc(a, a, mod);
		b >>= 1;
	}
	return ans;

}
inline ll read() {
	ll s = 0; bool w = true;
	char c = getchar();
	for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
	for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
	return w ? s : -s;
}
inline void fp(ll n) {
	if (n < 0)  putchar('-'), n = -n;
	if (n > 9) fp(n / 10);
	putchar(n % 10 + 48);
}
const int N = 3e6 + 10;
ll inv[N], s[N],a[N];
void solve() {
	ll n = read();
	
	ll p = read();
	inv[1] = 1;
	for (int i = 2; i <= n; ++i)inv[i] = ((p - p / i) % p * inv[p % i] % p) % p;//不能调用快速乘，会超时
	for (int i = 1; i <= n; ++i)fpn(inv[i]);
}
signed main()
{
	
	ll T = 1;
	//   T = read();
	while (T--) {

		solve();
	}
	return 0;
}
/**
*　　┏┓  WQJ ┏┓+ +
*　┏┛┻━━━┛┻┓ +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　 　　　    ┣┓>>>
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/

第三种，利用exgcd（二元函数的解求逆元），gcd(a,p）=1,p不为素数也可

#include<bits/stdc++.h>
#include<unordered_map>
#include<math.h>
#include<forward_list>
#include<unordered_set>
using namespace std;
#pragma warning(disable:4996)
const double pi = acos(-1);
#define pdd pair<double,double>
#define re register
#define ll long long
#define endl '\n'
#define pii pair<ll,ll>
#define IOS std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ull unsigned ll
#define eb emplace_back
#define pb push_back
#define rad read
#define mp make_pair
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
#define fpn(x) fp(x),putchar('\n');
ll mod = 80112002;
inline ll  gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
inline ll exgcd(ll a, ll b, ll& x, ll& y) {
	//ax+by=1; 返回值为a,b最大公约数
	if (!b) {
		x = 1, y = 0;
		return a;
	}
	ll g= exgcd(b, a % b, y, x);
	y -= (a / b) * x;
	return g;
}
inline ll ksc(ll a, ll b, ll mod) {

	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b, ll mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)
			ans = ksc(ans, a, mod);
		a = ksc(a, a, mod);
		b >>= 1;
	}
	return ans;

}
inline ll read() {
	ll s = 0; bool w = true;
	char c = getchar();
	for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
	for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
	return w ? s : -s;
}
inline void fp(ll n) {
	if (n < 0)  putchar('-'), n = -n;
	if (n > 9) fp(n / 10);
	putchar(n % 10 + 48);
}
const int N = 3e6 + 10;
ll inv[N], s[N],a[N];
void solve() {
	//求1-n的逆元
	ll n = read();
	ll p = read();
	inv[1] = 1;
	for (int i = 2; i <= n; ++i) {
		ll x = 0, y = 0;
		exgcd(i, p, x, y);//ix+py=gcd(i,p)（%p）  此处x=i的-1次方（即i的逆元）
		inv[i] = (x+p)%p;   //确保它为整数
	}
	for (int i = 1; i <= n; ++i)fpn(inv[i]);
}
signed main()
{
	
	ll T = 1;
	//   T = read();
	while (T--) {

		solve();
	}
	return 0;
}
/**
*　　┏┓  WQJ ┏┓+ +
*　┏┛┻━━━┛┻┓ +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　 　　　    ┣┓>>>
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/