这里只有模板，并不作讲解，仅为路过的各位做一个参考以及用做自己复习的资料，转载注明出处。
逆元
费马小定理
/*Copyright: Copyright (c) 2018
*Created on 2018-11-07
*Author: 十甫
*Version 1.0
*Title: 逆元-费马小定理
*Time: 2 mins
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
inline ll mul(ll a, ll b, ll mod) {
ll res = 1;
while(b) {
if(b & 1) res = ((res % mod) * (a % mod));
a = ((a % mod) * (a % mod));
b /= 2;
}
return res;
}
int main() {
ll n, k;
scanf("%lld%lld", &n, &k);
printf("%lld
", mul(n, k - 2, k));
return 0;
}EX-GCD
/*Copyright: Copyright (c) 2018
*Created on 2018-11-07
*Author: 十甫
*Version 1.0
*Title: 逆元-EX-GCD
*Time: 2 mins
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
inline void exgcd(ll a, ll b, ll &x, ll &y, ll mod) {
if(!b) {
x = 1, y = 0;
return;
}
exgcd(b, a % b, y, x, mod);
y -= (a / b) * x;
y = (y + mod) % mod;
}
int main() {
ll n, k;
scanf("%lld%lld", &n, &k);
ll a, b;
exgcd(n, k, a, b, k);
printf("%lld
", a);
return 0;
}线性递推
/*Copyright: Copyright (c) 2018
*Created on 2018-11-07
*Author: 十甫
*Version 1.0
*Title: 逆元-线性递推
*Time: inf mins
*/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int size = 10005;
int inv[size];
inline void make(int n, int p) {
inv[1] = 1;
for(int i = 2;i <= n;i++) {
inv[i] = inv[p % i] * (p - p / i) % p;
}
}
int main() {
int n, p;
scanf("%d%d", &n, &p);
make(n, p);
printf("%d
", inv[n]);
return 0;
}