标签:prime get stack url otherwise ted tom ota medium
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20711 Accepted Submission(s): 6500
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <list> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d\n", a) #define plld(a) printf("%lld\n", a) #define pc(a) printf("%c\n", a) #define ps(a) printf("%s\n", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 1100000, INF = 0x7fffffff; int prime[maxn + 10], phi[maxn + 10]; bool vis[maxn + 10]; int n, ans; void getphi() { ans = 0; phi[1] = 1; for(int i = 2; i <= maxn; i++) { if(!vis[i]) { prime[++ans] = i; phi[i] = i - 1; } for(int j = 1; j <= ans; j++) { if(i * prime[j] > maxn) break; vis[i * prime[j]] = 1; if(i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; } else phi[i * prime[j]] = phi[i] * (prime[j] - 1); } } } int q_pow(int a, int b, int mod) { int res = 1; while(b) { if(b & 1) res = res * a % mod; a = a * a % mod; b >>= 1; } return res; } vector<int> v; int main() { getphi(); while(cin >> n) { v.clear(); if(n % 2 == 0 || n == 1) { printf("2^? mod %d = 1\n", n); continue; } int res = phi[n]; for(int i = 2; i <= sqrt(phi[n] + 0.5); i++) { if(res % i == 0) v.push_back(i), v.push_back(res / i); } sort(v.begin(), v.end()); for(int i = 0; i < v.size(); i++) { if(q_pow(2, v[i], n) == 1) { res = v[i]; break; } } printf("2^%d mod %d = 1\n", res, n); } return 0; }
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20711 Accepted Submission(s): 6500
2^x mod n = 1 HDU - 1395(欧拉定理 原根)
标签:prime get stack url otherwise ted tom ota medium
原文地址:https://www.cnblogs.com/WTSRUVF/p/10800970.html