hdu 5187(高精度快速幂)

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1867    Accepted Submission(s): 596

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.

 

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)

 

Output

For each test case, output a single line indicating the answer.

 

Sample Input

2 233
3 5

 

Sample Output

2
1

Hint

In 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

BestCoder Round #33

 

题意:由 1-n n个数组成的序列，有多少种组合方式满足 a[1] - a[i] 单调递增/单调递减, a[i] - a[n] 单调递增/单调递减.

题解:如果想要a[1] - a[i] 单调递增,a[i] - a[n]单调递减,那么就只能将最大的n放到第 i 位置，那么当n的左边的排列确定了,那么其右边的排列也确定了,所以总共有C(0,n-1)+C(2,n-1)+...+C(n-2,n-1)+C(n-1,n-1) = 2^(n-1)种放法,反之就是将最小的1放到第i位置满足a[1]-a[i]单调递减,a[i]-a[n]单调递增了,情况数一样,但是由于单调递增和单调递减算了两次,所以要减2.答案为 2^n-2 ，由于n太大,所以才用了高精度快速幂(快速幂+快速乘法).

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
LL n,m;

LL modular_multi(LL a, LL b, LL c)  /// a * b % c
{
    LL res, temp;
    res = , temp = a % c;
    while (b)
    {
        if (b & )
        {
            res += temp;
            if (res >= c)
            {
                res -= c;
            }
        }
        temp <<= ;
        if (temp >= c)
        {
            temp -= c;
        }
        b >>= ;
    }
    return res;
}
LL modular_exp(LL a, LL b, LL c)
{
    LL res, temp;
    res =  % c, temp = a % c;
    while (b)
    {
        if (b & )
        {
            res = modular_multi(res, temp, c);
        }
        temp = modular_multi(temp, temp, c);
        b >>= ;
    }
    return res;
}
int main()
{
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        LL ans = modular_exp(,n,m);
        printf("%lld\n",(ans-+m)%m);
    }
    return ;
}