题意
设A(n) = n个1,问有多少对i,j使得\(A(i^j)\equiv0(modp)\)
题解
\(A(n) = \frac{10^n-1}{9}\)
当9与p互质时\(\frac{10^n-1}{9}\%p = (10^n-1)\cdot inv[9] \% p\)
移动项得到\(10^n\equiv1(modp)\)
由欧拉定理当\(gcd(10,p) = 1\)时\(10^{\varphi(p)}\equiv1(modp)\)
那么只要找到最小的d使得\(10^d\equiv1(modp)\)
问题就转化成求有多少对i,j使得\(i^j\equiv0(modp)\)
求d只需要枚举\(\varphi(p)\)的因子就好了
对d分解\(d = p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}\)
固定j，要使\(i^j\)是d的倍数，那么i一定是\(p_1^{\lceil\frac{k_1}{j}\rceil}p_2^{\lceil\frac{k_2}{j}\rceil}\cdots p_n^{\lceil\frac{k_n}{j}\rceil}\)的倍数
设\(g_j = p_1^{\lceil\frac{k_1}{j}\rceil}p_2^{\lceil\frac{k_2}{j}\rceil}\cdots p_n^{\lceil\frac{k_n}{j}\rceil}\),答案就是\(\sum_{j=1}^mg_j\),因为\(k_i\)不会超过30，
当j大于30时的\(g_j\)都一样就不用重复计算了
还有一个问题，当p=3时，因为9与3不互质,inv[9]不存在，式子\(\frac{10^n-1}{9}\%p \Longleftrightarrow (10^n-1)\cdot inv[9] \% p\)
就不成立，需要特判，此时d取3
代码
#include
using namespace std;
const int mx = 3e5+10;
typedef long long ll;
ll pow_mod(ll a, ll b, ll mod) {
ll ans = 1;
while (b > 0) {
if (b & 1) ans = ans * a % mod;
a = a * a % mod;
b /= 2;
}
return ans;
}
ll pow_mod(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b & 1) ans = ans * a;
a = a * a;
b /= 2;
}
return ans;
}
vector pp, k;
int main() {
int T;
scanf("%d", &T);
while (T--) {
ll p, n, m, d;
scanf("%lld%lld%lld", &p, &n, &m);
if (p == 2 || p == 5) {
printf("0\n");
continue;
}
d = p-1;
for (ll i = 1; i*i <= (p-1); i++) {
if ((p-1) % i == 0) {
if (pow_mod(10, i, p) == 1) {
d = min(d, i);
}
if (pow_mod(10, (p-1)/i, p) == 1) {
d = min(d, (p-1)/i);
}
}
}
if (p == 3) d = 3;
pp.clear(); k.clear();
ll ans = 0;
for (ll i = 2; i*i <= d; i++) {
if (d % i == 0) {
int tmp = 0;
while (d % i == 0) {
tmp++;
d /= i;
}
k.push_back(tmp);
pp.push_back(i);
}
}
if (d > 1) pp.push_back(d), k.push_back(1);
ll tmp = 1;
for (int i = 1; i <= min(30LL, m); i++) {
tmp = 1;
for (int j = 0; j < pp.size(); j++) {
ll b = k[j] / i;
if (k[j] % i != 0) b++;
tmp *= pow_mod(pp[j], b);
}
ans += n / tmp;
}
if (m > 30) ans += n / tmp * (m-30);
printf("%lld\n", ans);
}
return 0;
}