一个数在modP意义下有逆元当且仅当这个数与P互质，否则无论成什么数都不能成一
且逆元有且只有一个
所以只要统计P之前的（不含P）与P互质的数的个数 s
但因为有可能算两次，所以找出所有x*x=1(mod P) 个数是t
答案就是（s+t）/2
而这个P之前与P互质的数的个数就是欧拉函数
分解质因数
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <vector> 11 using namespace std; 12 typedef long long ll; 13 typedef unsigned int uint; 14 typedef unsigned long long ull; 15 typedef pair<int, int> PII; 16 #define fi first 17 #define se second 18 #define MP make_pair 19 20 ll read() 21 { 22 ll v = 0, f = 1; 23 char c = getchar(); 24 while (c < 48 || 57 < c) {if (c == '-') f = -1; c = getchar();} 25 while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar(); 26 return v * f; 27 } 28 29 int main() 30 { 31 freopen("count.in", "r", stdin); 32 freopen("count.out", "w", stdout); 33 ll n = read(); 34 ll _n = n, phi = n; 35 for (ll i = 2; i * i <= _n; i++) 36 if (_n % i == 0) 37 { 38 phi = phi / i * (i - 1); 39 while (_n % i == 0) 40 _n /= i; 41 } 42 if (_n > 1) 43 phi = phi / _n * (_n - 1); 44 ll s = 0; 45 for (ll i = 1; i < n; i++) 46 if (i * i % n == 1) 47 s++; 48 printf("%lld ", (phi + s) / 2); 49 }
欧拉定理求逆元
