## Multiplicative inverse

### 1. Fermat's Little Theorem

If \$ a, P \$ prime, then \$ a ^ {p-1} ≡ 1 (mod p) \$
and \$ a known equation by the inverse of x ≡ 1 (mod p) \$ , to give \$ a x≡ A ^ { p-1} (mod p) \$
so as inverse \$ X \$ \$ a ^ {p-2} mod p \$, fast power to solve.

``````    #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

#define LL long long

LL n,p;

inline LL fast_pow(LL a,LL b,LL p) {
LL ans = 1;
a %= p;
while(b) {
if(b&1) ans = ans * a % p;
a = a * a % p;
b >>= 1;
}
ans %= p;
return ans;
}

int main() {
scanf("%lld%lld",&n,&p);
for(int i = 1 ; i <= n ; i++)
printf("%lld \n",fast_pow(i,p-2,p));
return 0;
}``````

### 2. Linear inversing

This is not evidence.
But will not permit problem is not large, can be directly used as a conclusion.
\$ Inv [i] = p - \ frac {p} {i} * inv [p mod i] mod p \$
Note, however \$ INV [. 1] =. 1, INV [0] = tan90 ^ O = 0 \$
\$ for \$ 2 cycle from the start, the time complexity \$ O (n) \$.

``````

using namespace std;

#define LL long long
const int N = 3e6 + 10;

LL n,p,inv[N];

int main() {
scanf("%lld%lld",&n,&p);
inv[1] = 1;inv[0] = 0;
for(int i = 2; i <= n ; i++)
inv[i] = p - (p / i) * inv[p % i] % p;
for(int i = 1 ; i <= n ; i++)
printf("%lld \n",inv[i]);
return 0;
}
``````

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Origin www.cnblogs.com/Repulser/p/11207140.html
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