hdu 5187 zhx's contest （排列组合+快速乘法+快速加法）

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1308    Accepted Submission(s): 421

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

如果n=1，答案是1，否则答案是2n−2。

证明：ai肯定是最小的或者最大的。考虑ai前面的数，如果它们的位置定了的话，后面剩下的数的排列顺序是唯一的。

即随着ai位置的变动，取C(n-1,0)+C(n-1,1)+...+C(n-1,n-1)=2^(n-1);

那么ai是最小或者最大分别有2n−1种情况，而整个序列单调增或者单调减的情况被算了2次，所以要减2。

要注意的一点是因为p>231，所以要用快速乘法。用法与快速幂相同。如果直接乘会超过long long范围，从而wa掉。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
using namespace std;
#define N 50005
#define ll __int64
const int inf=0x7fffffff;
ll p;
ll mul(ll a,ll b) //快速加法
{
    ll ans=0;
    while(b)
    {
        if(b&1)
            ans=(ans+a)%p;
        a=(a+a)%p;
        b>>=1;
    }
    return ans;
}
ll pow(ll a,ll b)   //快速乘法
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=mul(ans,a);
        a=mul(a,a);
        b>>=1;
    }
    return ans;
}
int main()
{
    ll n,ans;
    while(scanf("%I64d%I64d",&n,&p)!=-1)
    {
        if(n==1)
            printf("%I64d\n",1%p);
        else
        {
            ans=pow(2ll,n)-2;
            printf("%I64d\n",(ans+p)%p);      //去掉2可能为负数
        }
    }
    return 0;
}