HDU-4335-What is n? -Number Theory

Number theory. Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4335

Given bpm, several N conditions are required.

Proof see http://hi.baidu.com/aekdycoin/item/e493adc9a7c0870bad092fd9

The number of B, P, and m in the given question 0 <= n <= m must be n ^ (N !) Partition B (mod P), and the bad thing is that we didn't mention any special properties of the number here. Given is only limited to a positive integer and P> 0.

With the Euler's theorem, we know that N ^ (PHI (p) has 1 (mod P), but gcd (n, p) = 1 is required here. Obviously, the question does not have such data, if the given question is N or P, then we can know
N ^ (x) mod P has a circular section. This circular section is n ^ (PHI (p). If N and P are not of mutual quality, so we can prove that this cycle Section T | PHI (P), so we can still choose PHI (p) as the cycle section, so
Next we can divide the calculation into four parts.

Part_1: calculate n first! <PHI (p) has eight points at most, and the for loop is directly violent.

Part_2: if there are still intervals left, calculate PHI (p) <= n! <S !, S = min {x | X! MoD PHI (p) = 0} // n at this time! The power already has a circular section, but n is still changing.

Part_3: The next interval is n! % PHI (p) = 0, which indicates that it is completely dependent on the Phi (p) value, so as long as the cyclic nature of N is taken into account, simply count the number of occurrences in a loop section of N.

Part_4: Count the remaining n.

# Include <cstdlib> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # define maxn 100000 using namespace STD; typedef unsigned long int int64; int P [maxn + 5], PM [10000], Phi [maxn + 5], idx =-1, Mod, B; int FAC [maxn + 5]; int64 m, ans; void getprime () {for (INT I = 2; I <= maxn; ++ I) {If (! P [I]) {PM [++ idx] = I;} For (Int J = 0; PM [J] * I <= maxn; ++ J) {P [PM [J] * I] = 1; if (I % PM [J] = 0) {break ;}}} /* printf ("% d \ n", idx); For (INT I = 0; I <= 100; ++ I) {printf ("% d ", PM [I]);} */} void Eular () {Phi [1] = 1; for (INT I = 2; I <= maxn; ++ I) {// printf ("Phi [% d] = % d \ n", I-1, Phi [I-1]); // getchar (); If (! P [I]) {Phi [I] = I-1; continue;} For (Int J = 0; PM [J] * PM [J] <= I; ++ J) {if (I % PM [J] = 0) {If (I/PM [J] % PM [J] = 0) {Phi [I] = PM [J] * Phi [I/PM [J];} else {Phi [I] = Phi [PM [J] * Phi [I/PM [J];} break ;}}} bool _ POW (int64, int64 B) {int64 ret = 1; while (B) {If (B & 1) {RET * = A; RET % = MOD;} A * =; A % = MOD; B >>=1;} return ret = B;} void deal () {int Lim =-1, PTR; For (PTR = 0; PTR <= m; ++ PTR) {// the first segment, less than a part of Phi [mod] If (! PTR) {FAC [PTR] = 1;} else {FAC [PTR] = FAC [ptr-1] * PTR;} If (FAC [PTR]> = Phi [mod]) {break;} If (_ POW (PTR, FAC [PTR]) {++ ans ;}} if (PTR> m) {return ;} // If M is exceeded, for (INT I = PTR; I <= m; ++ I) {If (! I) {FAC [I] = 1% Phi [mod];} else {FAC [I] = (FAC [I-1] * I) % Phi [mod];} if (FAC [I] = 0) {// If This vertex is found Lim = I; break;} If (_ POW (I, FAC [I] + Phi [mod]) {++ ans ;}}if (lim =-1) {return ;} // if no vertex int T = 0 that is multiples of Phi [mod] is found; If (m-lim + 1)> = mod) {// Lim this point is not calculated, so the length is M-LIM + 1, Here determine whether the remaining is greater than mod for (INT I = lim; I <Lim + MOD; ++ I) {If (_ POW (I % mod, Phi [mod]) {// because it is already a multiple of Phi [mod, therefore, the Phi [mod] index is directly substituted into ++ t;} ans + = (m-lim + 1)/(1llu * mod) * t ;} int left = (M-LIM + 1) % MOD; // residual remainder for (INT I = lim; I <Lim + left; ++ I) {If (_ POW (I % mod, Phi [mod]) {++ ans ;}} int main () {getprime (); Eular (); int t, CA = 0; scanf ("% d", & T); While (t --) {ans = 0; scanf ("% d % i64u", & B, & mod, & M); If (M = 18446731673709551615llu & mod = 1 & B = 0) {// because B = 0, P = 1, therefore, you can obtain m + 1 directly. Therefore, printf ("case # % d: 18446744073709551616 \ n", ++ CA);} else {deal (); printf ("case # % d: % i64u \ n", ++ ca, ANS) ;}} return 0 ;}