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 Description
As 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.
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.
Input
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)
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)
Output
For each test case, output a single line indicating the answer.
Sample Input
2 233
3 5
Sample Output
2
1HintIn 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
Source
思路：由题意能够求出答案为(2^n-2)%p
可是n。p都是LL型的，高速幂的时候会爆LL，所以这里要用到高速乘法，高速乘法事实上和高速幂差点儿相同。就是把乘号改为加号
注意：当n为1时。要输出1，而当p为1时要输出0。
AC代码：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std; LL n, p; LL multi(LL a, LL b) { //高速乘法。事实上和高速幂差点儿相同
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) { //高速幂
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;
}