题目链接：uva 11256 - Repetitive Multiple

题目大意：给定一个数n，要求找到最小的k，使得k?n为题目中定义的重复数字.

解题思路：枚举k?n的循环节长度，比如当前枚举为2，那么一次判断u=1001，1001001，1001001001 ...，取d = gcd(n,u), 那么k = u / d, a = n / d （因为n?k=u?a）并且保证a的长度为2，所以k和a要同时扩大相应倍数。枚举过程中为何k。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;

const ll INF = 1e18+1;
ll N, ten[20];

void init () {
    ten[0] = 1;
    for (int i = 1; i < 20; i++)
        ten[i] = ten[i-1] * 10;
}

inline ll gcd (ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}

inline int tenbit (ll n) {
    int cnt = 0;
    while (n) {
        n /= 10;
        cnt++;
    }
    return cnt;
}

ll solve () {
    ll ans = INF;
    int bit = tenbit(N);

    if (bit == 1)
        return 11L;

    for (int i = 1; i <= bit; i++) {
        ll u = ten[i] + 1;

        while (true) {
            ll d = gcd(N, u);

            ll a = N / d;
            ll k = u / d;


            if (a < ten[i]) {

                if (a < ten[i-1])
                    a = ten[i-1] / a + (ten[i-1] % a ? 1 : 0);
                else
                    a = 1;

                if (a * k < ans)
                    ans = k * a;
            }

            if (INF / u < ten[i])
                break;

            u = u * ten[i] + 1;
        }
    }
    return ans;
}

int main () {
    init();

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%lld", &N);
        while (N >= 1e9);
        printf("%lld\n", solve() * N);
    }
    return 0;
}