HDU, hdukmp

HDU, hdukmp

Zhx's contest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 448 Accepted Submission (s): 147


Problem DescriptionAs one of the most powerful brushes, zhx is required to give his juniors N Problems.
Zhx thinks Ith Problem's difficulty is I . He wants to arrange these problems in a beautiful way.
Zhx defines a sequence {Ai} Beautiful if there is I That matches two rules below:
1: A1.. ai Are monotone decreasing or monotone increasing.
2: Ai .. Are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there if the sequence of the problems 'difficulty is beautiful.
Zhx knows that the answer may be very huge, and you only need to tell him the answer module P .
InputMultiply test cases (less 1000 ). Seek EOF As the end of the file.
For each case, there are two integers N And P Separated by a space in a line .( 1 ≤ n, p ≤ 1018 )
OutputFor each test case, output a single line indicating the answer.
 
Sample Input

2 2333 5

 
Sample Output

21HintIn the first case, both sequence {1, 2} and {2, 1} are legal.In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1 

 
SourceBestCoder Round #33

The answer is (2 ^ N-2) % p.

However, n and p are all LL-type, and the LL will be exploded when the power is saved. Therefore, fast multiplication is used here. Fast Multiplication is similar to the power of Fast Multiplication, that is, changing the multiplication number to the plus sign.

Note: When n is 1, 1 is output, and when p is 1, 0 is output;

AC code:

# Include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # define LL long using namespace std; LL n, p; LL multi (LL, LL B) {// fast multiplication, which is similar to the Fast Power LL ret = 0; while (B) {if (B & 1) ret = (ret + a) % p; a = (a + a) % p; B >>= 1;} return ret;} LL powmod (LL a, LL B) {// fast power LL ret = 1; while (B) {if (B & 1) ret = multi (ret, a) % p; a = multi (a, a) % p; B >>= 1 ;} return ret;} int main () {while (cin> n> p) {if (p = 1) {cout <0 <endl ;} else if (n = 1) {cout <1 <endl;} else {LL ans = powmod (2, n)-2; if (ans <0) ans + = p; cout <ans <endl ;}} return 0 ;}