hdu 5187(快速幂+快速乘法)

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

[align=left]Problem Description[/align]
As one of the most powerful brushes, zhx is required to give his juniorsn
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 modulep.
 

[align=left]Input[/align]
Multiply 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)
 

[align=left]Output[/align]
For each test case, output a single line indicating the answer.

 

[align=left]Sample Input[/align]

2 233
3 5

 

[align=left]Sample Output[/align]

2
1

HintIn 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

 
思路：首先是全部递减或全部递增各一种；另外就是满足上列两个条件的情况了，要想满足条件（1）那就只能把最大的n放在i位置，共有C（1，n-1）+C（2，n-1）+。。。+C（n-2，n-1）即2^(n-1)-2；条件（2）与（1）相同，所以共有(2^(n-1)-2)*2+2=2^n-2.

#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，其实和快速幂差不多
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) {	//快速幂
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;
}