uva 11256 - Repetitive Multiple(gcd+暴力)

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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;}

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