acwing:https://www.acwing.com/problem/content/878/
求逆元
#include <bits/stdc++.h>
using namespace std;
#define LL long long
LL n, a, p;
LL qp(LL a, LL k, LL p){
LL ans = 1;
while (k){
if (k & 1) ans = ans * a % p;
k >>= 1;
a = a * a % p;
}
return ans;
}
//1.快速幂求逆元
LL inv(LL x){
return qp(x, mod - 2, mod);
}
//2.扩展欧几里得求逆元
//LL inv(LL a){
// LL x, y;
// exgcd(a, mod, x, y);
// x = (x % mod + mod) % mod;
// return x;
//}
//3.线性递推求逆元
//inv[1] = 1;
//for (int i = 2; i <= n; i ++ )
// inv[i] = (p - p / i) * inv[p % i] % p;
int main(){
ios::sync_with_stdio(false);cin.tie(0);
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> a >> p;
if (a % p == 0) cout << "impossible\n";
else cout << inv(a, p) << "\n";
}
return 0;
}
luogu:https://www.luogu.com.cn/problem/P3811
给定 \(n\),\(p\) 求 1 ~ \(n\) 中所有整数在模 \(p\) 意义下的乘法逆元。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 3e6 + 10;
LL n, p, inv[N];
int main(){
cin >> n >> p;
inv[1] = 1;
cout << inv[1] << "\n";
for (int i = 2; i <= n; i ++ ){
inv[i] = (p - p / i) * inv[p % i] % p;
cout << inv[i] << "\n";
}
return 0;
}
luogu:https://www.luogu.com.cn/problem/P5431
给定 \(n\) 个正整数 \(a_i\),求它们在模 \(p\) 意义下的乘法逆元,给定常数 \(k\),输出 \(\sum_{i = 1}^n \frac{k^i}{a_i}\),答案对 \(p\) 取模。
题目卡常,要快读。
思路:
定义 \(s = \prod_{i = 1}^n\)。原式就转化为 \(\frac{k^i * \frac{s}{a_i}}{s}\)。
预处理一下前缀积和后缀积,就可以求了。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 5e6 + 10;
LL ans, n, p, k, a[N], pre[N], suf[N];
inline LL read(){
LL s = 0, w = 1;
char ch = getchar();
while ( ch < '0' || ch > '9') { if (ch == '-') w = -1; ch = getchar();}
while ( ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s * w;
}
LL qp(LL a, LL k, LL p){
LL ans = 1;
while (k){
if (k & 1) ans = ans * a % p;
k >>= 1;
a = a * a % p;
}
return ans;
}
LL inv(LL x){//快速幂求逆元
return qp(x, p - 2, p);
}
int main(){
n = read(), p = read(), k = read();
pre[0] = 1, suf[n + 1] = 1;
for (int i = 1; i <= n; i ++ ){
a[i] = read();
pre[i] = pre[i - 1] * a[i] % p;
}
for (int i = n; i >= 1; i -- )
suf[i] = suf[i + 1] * a[i] % p;
for (LL i = 1, j = k; i <= n; i ++, j = j * k % p )
ans = (ans + ( (j * pre[i - 1] % p) * suf[i + 1]) % p ) % p;
cout << ( ans * inv(pre[n]) ) % p << "\n";
return 0;
}
应用:
2021 CCPC 威海站:https://www.cnblogs.com/Hamine/p/16029602.html