HDU 2462 The Luckiest number

题意很好理解，就不多说了。

可以很容易想到是找满足 10^x = 1 (mod 9 * L / gcd(L, 8))中最小的x，且x>0。

对于要求的数N，N = 8 *( 11.....1 )= 8 * ( 10 ^ x - 1 ) / 9。

要求N = 0 ( mod L ) => 8 * ( 10 ^ x - 1 ) / 9 = 0 (mod L) => 10 ^ x - 1 = 0 (mod 9L / gcd(L,8)) => 10 ^ x = 1 (mod 9L / gcd(L,8))。

设k = 9L / gcd(L,8)，则即是求满足10 ^ x = 1 (mod k)的最小x。

一看到这个姿势第一反应就是费马-欧拉定理：

其中a和n为正整数且互素。

所以若10与k不互素，即不满足欧拉定理，此时无解。

剩下的就是求10 ^ phi(k) = 1 (mod k)。蓝儿phi(k)不一定是最小的。既然phi(k)和10互素，phi(k)的因子也一定和10互素。所以，枚举因子，找到最小的得出答案。

蓝儿题目数据较大，所以快速幂运算时不能直接乘，要用到乘法转化，提前取模，以免爆LL。

代码：

// Header.#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <set>#include <map>using namespace std;// Macrotypedef long long LL;#define TIME cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;#define IN freopen("/Users/apple/input.txt", "r", stdin);#define OUT freopen("/Users/apple/out.txt", "w", stdout);#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define ALL(v) v.begin(), v.end()#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)#define put(a) printf("%d\n", a)#define ss(a) scanf("%s", a)#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define VI vector<int>#define pb push_backconst int inf = 0x3f3f3f3f, N = 5e4 + 5, MOD = 1e9 + 7;const LL INF = 1e18;// Macro endint T, cas = 0;LL n;// Impbool vis[N];vector<int>primes;vector<LL> pri;LL gcd(LL a, LL b) {return !b ? a : gcd(b, a % b);}void Init_Primes() {mem(vis, 0);int m = sqrt(N + 0.5);FOR(i, 2, m) if(!vis[i])for(int j = i * i; j <= N; j += i) vis[j] = 1; REP(i, 2, N) if(!vis[i])primes.pb(i);}LL Euler_Phi(LL n) {LL ret = n, m = (LL)sqrt(n + 0.5);FOR(i, 2, m) if(n % i == 0) {ret = ret / i * (i - 1);while(n % i == 0) n /= i;} if(n > 1) ret = ret / n * (n - 1);return ret;}void Phi_Table(int n, int * phi) {mem(phi, 0);phi[1] = 1;FOR(i, 2, n) {if(!phi[i]) {for(int j = i; j <= n; j += i) {if(!phi[j]) phi[j] = j;phi[j] = phi[j] / i * (i - 1);}}}}LL Mul(LL a, LL b, LL m) {LL ret = 0;while(b) {if(b & 1) ret = (ret + a) % m;a = a * 2 % m;b >>= 1;}return ret;}LL Pow(LL a, LL b, LL m) {LL ret = 1;while(b) {if(b & 1) ret = Mul(ret, a, m);a = Mul(a, a, m);b >>= 1;}return ret;}void init(LL n) {pri.clear();int sz = primes.size();rep(i, sz) while(n % (LL)primes[i] == 0) {n /= primes[i];pri.pb(primes[i]);} if(n > 1) pri.pb(n);}int main(){#ifdef LOCAL    IN // OUT#endifInit_Primes();    while(scanf("%lld", &n) != EOF, n) {    LL m = 9 * n / gcd(n, 8LL);    if(gcd(m, 10LL) != 1) {    printf("Case %d: %d\n", ++ cas, 0);    continue;    }    LL x = Euler_Phi(m);    init(x);    int sz = pri.size();    rep(i, sz) if(Pow(10LL, x / pri[i], m) == 1) x /= pri[i];    printf("Case %d: %lld\n", ++ cas, x);    }    return 0;}