洛谷P4884 多少个1？

(large{题目链接})
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(Large extbf{Soluiton: } large{题意可化为求最小的n满足sum ^{n-1}limits_{i=0}10^{i}equiv kleft(mod m ight)，可转化为dfrac {10^{n}-1}{9}equiv kleft(mod m ight)，观察这个式子，用 ext{BSGS}求即可})
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(Large extbf{Summary: } large{1.龟速乘可以防止爆long long。\2.膜一下tlx，比我快，空间还比 我小，orz。})
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(Large extbf{Code:})

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

ll k, m;
map <ll, ll> h; 

inline ll read() {
	ll x = 0;
	char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return x; 
}

inline ll mul(ll x, ll y, ll p) {
	ll z = (ld)x / p * y;
	ll ret = (ull)x * y - (ull)z * p;
	return (ret + p) % p;
}

int main() {
	k = read(), m = read();
	k = (mul(9, k, m) + 1) % m;
	int t = ceil(sqrt(m)); ll cur = 1;
	h[k % m] = 0;
	for (int i = 1; i <= t; ++i) cur = mul(cur, 10, m), h[mul(cur, k, m)] = i;
	ll ans = 1;
	for (int i = 1; i <= t; ++i) {
		ans = mul(ans, cur, m);
		if (h[ans]) { printf("%lld
", ((1ll * i * t - h[ans]) % m + m) % m); return 0; }
	}
}