Hdu4335 -- What is n? -- Number Theory

Question address: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4335

Question meaning:

Here are three numbers: B, P, M.

Ask if you meet 0 <= n <= m and n ^ n! Number of N in % P = B

Solution:

Main Research

A ^ X % C = a ^ (X % PHI (c) + PHI (c) % C where x> = PHI (c)

Phi (c) is the Euler's function.

Part 1 n! <PHI (c) Then you need to calculate n directly! Fortunately, Phi (c) is not very large.

Part 2 n! > = PHI (c) but n! % PHI (c )! = 0 this part also requires brute-force computation, but the modulo operation can be performed without a very large number.

Part 3 n! > = PHI (C) and n! % PHI (c) = 0 is converted to N ^ PHI (c) % C and then (N % C) ^ PHI (c) % C is a loop section with the length of C.

For details, see the code:

# Include <cstdio> # include <iostream> # include <cstring> # include <cmath> using namespace STD; # define ull unsigned long // calculate the Euler's function const int maxp = 100000 + 20; ull Phi [maxp]; void phi_table () {memset (PHI, 0, sizeof (PHI); Phi [1] = 1; for (INT I = 2; I <maxp; I ++) {If (! Phi [I]) {for (Int J = I; j <maxp; j ++ = I) {If (! Phi [J]) Phi [J] = J; Phi [J] = Phi [J]/I * (I-1); }}} ull euler_phi (ull P) {ull M = (ull) SQRT (p + 0.5); ull ans = P; For (ull I = 2; I <= m; I ++) {If (P % I = 0) {ans = ANS/I * (I-1); While (P % I = 0) P/= I ;}} if (p> 1) ans = ANS/P * (p-1); Return ans;} ull power (ull A, ull X, ull p) {ull ret = 1; A = A % P; while (x> 0) {If (X & 1) ret = RET * A % P; A = A * A % P; X = X/2;} return ret;} int main () {// phi_table (); int t; int CA = 1; scanf ("% D", & T); While (t --) {ull B, P, M; CIN> B> P> m; // If (B = 0 & P = 1 & M = 18446731673709551615ull) {printf ("case # % d: 18446744073709551616 \ n", CA ++); continue;} ull _ Phi = euler_phi (p); ull ans = 0; if (B = 0) ans ++; ull sum = 1; // first look at n! % Phi! = 0, when n! <Phi, MOD is not 0 for (ull I = 1; I <_ Phi & I <= m; I ++) {sum = sum * I % _ Phi + _ Phi; If (Power (I, sum, P) = B) ans ++ ;} // and then from Phi to m, but we noticed n next! % Phi = 0, then the equation is I ^ Phi % P = B // then (I + KP) ^ Phi % P = B, from the beginning of I to m, we can see several P lengths for (ull I = _ Phi; I <_ Phi + P & I <= m; I ++) {If (Power (I, _ Phi, P) = B) {ans + = (M-I)/p + 1 ;}} cout <"case #" <ca ++ <":" <ans <Endl;} return 0 ;}