Poj-3696 The Luckiest number(数论)

最新推荐文章于 2020-07-04 16:44:24 发布

原创最新推荐文章于 2020-07-04 16:44:24 发布 · 289 阅读

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本文介绍了一个有趣的问题：如何找出由数字8组成的、最小的且能被特定整数整除的正整数。通过数学推导和算法实现，文章提供了一种有效的方法来解决这个问题。

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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

Sample Output

Case 1: 1
Case 2: 2
Case 3: 0

分析: WA到怀疑人生,最后找了网上的代码发现是取模时溢出了...

88888....可以表示为(10^k-1)/9*8,现在我们想让(10^k-1)/9*8 = L*m,也就是8*(10^k-1) = 9*L*m,去掉8和L的公约数,整个式子可以化简为p*(10^k-1) = m*q,其中p q互质,那么我们就是要找到最小的k使得

(10^k-1) % q = 0,也就是10^k  ≡ 1 (% q),如果q和10不互质则无解,否则我们可以先用欧拉函数求出一个解,然后枚举这个解的所有因数来得到最小的解.

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
using namespace std;
typedef long long  ll;
int Time;
ll l;
ll muti(ll a,ll b,ll MOD)
{
    ll ans = 0;
    while(b)
    {
        if(b & 1) ans = (ans + a) % MOD;
        a = (a<<1) % MOD;
        b>>=1;
    }
    return ans;
}
ll ksm(ll x,ll y,ll MOD)
{
    ll ans = 1;
    while(y)
    {
        if(y & 1) ans = muti(ans,x,MOD);
        x = muti(x,x,MOD);
        y >>= 1;
    }
    return ans;
}
ll euler_phi(ll n)
{
    ll ans = n;
    for(ll i = 2;i*i <= n;i++)
     if(n % i == 0)
     {
        ans = ans / i * (i-1);
        while(n % i == 0) n/=i;
     }
    if(n > 1) ans = ans / n * (n-1);
    return ans;
}
ll log_mod(ll a,ll b)
{
    ll temp = euler_phi(b),now = temp;
 //   cout<<temp<<endl;
    if(now == 1) return now;
    for(ll i = 1;i*i <= temp;i++)
     if(temp % i == 0)
     {
        if(ksm(a,i,b) == 1)
        {
            now = min(now,i);
            break;
        }
        if(ksm(a,temp/i,b) == 1) now = min(now,temp/i);
     }
    return now;
}
int main()
{
	while(cin>>l && l)
	{
		ll p = 9*l/__gcd(l,8ll);
        if(__gcd(p,10ll) != 1ll)  printf("Case %d: 0\n",++Time);
		else printf("Case %d: %I64d\n",++Time,log_mod(10,p));
	}
}
/*/
999999999
2000000000
/*/