[HDOJ1395]2^x mod n = 1（欧拉函数）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1395

题意：给出n，找2^x mod n = 1的最小的x。

欧拉函数是费马小定理的一个特殊应用：2^phi(x)=1 (mod x)，当且仅当x为素数的时候式子成立，因为x为素数时，比x小且与x互质的数的个数为x-1，即满足费马小定理2^(x-1)=1(mod x)。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 5500;
LL n;
LL geteular(LL x) {
LL ret = 1;
LL i;
for(i = 1; i < maxn; i++) {
ret *= 2;
ret %= x;
if(ret == 1) break;
}
return i;
}

int main() {
// FRead();
while(cin >> n) {
if(n == 1 || n % 2 == 0) {
printf("2^? mod %I64d = 1\n", n);
continue;
}
cout << "2^" << geteular(n) << " mod " << n << " = 1" << endl;
}
RT 0;
}