HDU 5187 zhx's contest（思维，快速幂，快速乘）

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2541    Accepted Submission(s): 783

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 2333 5

 

Sample Output

21
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

 

Recommend

hujie

这题我们可以借助DP的思想推出公式， [ 1 ~ N ] 所有满足条件的排列就等于 [ 1 ~ N -1 ] 所有满足条件的排列插进去一个 N，其中单调递增和单调递减排列我们可以将N插进三个位置          例如：  原排列为 [ 1 , 2 , 3 ] ，我们可以将4插入将其修改成 [ 1 , 2 , 3 , 4] 、[ 1 , 2 , 4 , 3]、[ 4 , 1 , 2, 3 ]，而增减序列和减增序列只能将N插入两个位置  （ [ 1 , 3 , 2 ]  改为 [ 1 , 3 , 4 ,2]、[ 1 , 4 , 3 , 2] ），显而易见状态转移方程为dp[ i ] = dp[ i - 1 ] * 2 + 2

然而这题的数据范围太大，并不允许我们dp求解，但我们可以借助dp打表推公式。。。。

显而易见公式为 2 ^ n -2，然而这题数据范围太大，直接用快速幂会爆 long long ，所以还需加一步快速乘。

PS：数学不好真心伤不起啊。。忙活半天才把公式找出来。。。。。。。

#include<bits/stdc++.h>using namespace std;long long n,Mod;long long mul(long long a,long long b){    long long res=0;    while(b)    {        if(b&1) res=(a+res)%Mod;        b>>=1;        a=(a+a)%Mod;    }    return res;}long long xx(long long a,long long n){    long long ret=1;    long long tmp=a%Mod;    while(n)    {        if(n&1)            ret=mul(ret,tmp)%Mod;        tmp=mul(tmp,tmp)%Mod;        n>>=1;    }    return ret;}int main(){    while(scanf("%I64d%I64d",&n,&Mod)==2)    {        printf("%I64d\n",(xx(2,n)%Mod-2+Mod)%Mod);  ///2^n-2可能为负数，所以要加一个Mod    }}