HDU 2256 Problem of Precision number theory matrix fast power, hdu2256

HDU 2256 Problem of Precision number theory matrix fast power, hdu2256

You must obtain the integer (√ 2 + √ 3) of 2n and then mod 1024.

(√ 2 + √ 3) 2n = (5 + 2 √ 6) n

If the value is calculated directly, double is used to store the value. When n is large, the loss of precision will increase and the expected result cannot be obtained.

We found that (5 + 2 √ 6) n + (5-2 √ 6) n is an integer (the even power of 2 √ 6 is always offset by plus or minus) and (5-2 √ 6) n is less than 1. So we just need to find Sn-1. Ling

An = (5 + 2 √ 6) n; Bn = (5-2 √ 6) n.

Sn = An + Bn Sn is An integer.

Sn * (5 + 2 √ 6) + (5-2 √ 6) = Sn * 10

Sn * 10 = (5 + 2 √ 6) n + 1 + (5-2 √ 6) n + 1 + (5 + 2 √ 6) n-1 + (5-2 √ 6) n-1

Sn * 10 = Sn + 1 + Sn-1

Recurrence: Sn = 10 * Sn-1-Sn-2

Then convert to a matrix to quickly calculate the Sn

# Include <iostream> # include <cstdio> # include <cstring> using namespace std; const int Mod = 1024; const int N = 2; struct Mat {int mat [N] [N];} a; Mat Multiply (Mat a, Mat B) {Mat c; memset (c. mat, 0, sizeof (c. mat); for (int k = 0; k <2; ++ k) for (int I = 0; I <2; ++ I) if (. mat [I] [k]) for (int j = 0; j <2; ++ j) if (B. mat [k] [j]) c. mat [I] [j] = (c. mat [I] [j] +. mat [I] [k] * B. mat [k] [j]) % Mod; return c;} Mat QuickPower (Mat a, int k) {Mat c; memset (c. mat, 0, sizeof (c. mat); for (int I = 0; I <2; ++ I) c. mat [I] [I] = 1; for (; k> = 1) {if (k & 1) c = Multiply (c, ); a = Multiply (a, a);} return c;} void InitMat (Mat & A) {. mat [0] [0] = 10;. mat [0] [1] =-1;. mat [1] [0] = 1;. mat [1] [1] = 0;} int main () {// freopen ("in.txt", "r", stdin); int t; scanf ("% d", & t); while (t --) {int n; scanf ("% d", & n); if (n = 1) printf ("9 \ n"); else if (n = 2) printf ("97 \ n"); else {InitMat (a); a = QuickPower (, n-2); int ans = (. mat [0] [0] * 98 +. mat [0] [1] * 10-1) % 1024; // we calculate S [n]-1 while (ans <0) ans + = 1024; printf ("% d \ n", ans) ;}} return 0 ;}