POJ 3696 The Luckiest Number 欧拉函数

是否存在n能整除的m使m各位都是8。

m=89(10x−1)=kn

8⋅10x−1gcd(8,n)=9⋅kngcd(8,n)

又gcd(8gcd(8,n),ngcd(8,n))=1

令q=9ngcd(8,n)

gcd(8gcd(8,n),q)=1

10x≡1modq

又aφ(p)≡1modp

的条件是gcd(a,p)=1

有gcd(10,q)=1时

10φ(q)≡1modq

不过φ(q)不是最小的，需要枚举φ(q)的因子并判断合法性。

似乎和原根求法有点像？

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
ll st[1000005], f;
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll phi(ll x) {
ll ans = 1, i, y = x;
for (i = 2; i * i <= y; ++i)
if (x % i == 0) {
x /= i, ans *= i - 1;
while (x % i == 0) x /= i, ans *= i;
}
if (x > 1) ans *= x - 1;
return ans;
}
ll quick_mul(ll a, ll b) {
ll ans = 0;
for (; b; b >>= 1, a = (a << 1) % f)
if (b & 1) ans = (ans + a) % f;
return ans;
}
ll quick_pow(ll a, ll b) {
ll ans = 1;
for (; b; b >>= 1, a = quick_mul(a, a))
if (b & 1) ans = quick_mul(ans, a);
return ans;
}
int main() {
int kase = 0;
ll x, m, i, sz;
while (scanf("%lld", &x) == 1 && x) {
f = 9 * x / gcd(8, 9 * x);
m = phi(f);
for (sz = 0, i = 1; i * i <= m; ++i) if (m % i == 0) {
st[sz++] = i;
if (i * i != m) st[sz++] = m / i;
}
sort(st, st + sz);
for (i = 0; i < sz; ++i) if (quick_pow(10, st[i]) % f == 1) {
printf("Case %d: %lld\n", ++kase, st[i]);
i = 2147483640;
}
if (i != 2147483641) printf("Case %d: 0\n", ++kase);
}
return 0;
}

The Luckiest number

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 4884 Accepted: 1313

Description

Chinese people think of ‘8’ as the lucky digit. Bob also likes digit ‘8’. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of only digit ‘8’.

Input

The input consists of multiple test cases. Each test case contains exactly one line containing L(1 ≤ L ≤ 2,000,000,000).

The last test case is followed by a line containing a zero.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the length of Bob’s luckiest number. If Bob can’t construct his luckiest number, print a zero.

Sample Input

8

11

16

0

Sample Output

Case 1: 1

Case 2: 2

Case 3: 0