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 the 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 an i that matches two rules below:
1: a1..ai are monotone decreasing or monotone increasing.
2: ai..an 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 than 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 

思路：由題意可以求出答案為(2^n-2)%p

但是n，p都是LL型的，快速冪的時候會爆LL，所以這裡要用到快速乘法，快速乘法其實和快速冪差不多，就是把乘號改為加號

注意：當n為1時，要輸出1，而當p為1時要輸出0；

AC代碼：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define LL long long using namespace std;LL n, p;LL multi(LL a, LL b) {//快速乘法，其實和快速冪差不多     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) {//快速冪     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;}