zhx's contest （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)

Total Submission(s): 352    Accepted Submission(s): 120

Problem Description

As one of the most powerful brushes, zhx is required to give his juniors 

n problems.

zhx thinks the 

i t h problem's difficulty is 

i . He wants to arrange these problems in a beautiful way.

zhx defines a sequence 

{ a i } beautiful if there is an 

i that matches two rules below:

1: 

a 1 . . a i are monotone decreasing or monotone increasing.

2: 

a i . . a n 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 

E O F 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 ≤ 10 18 )

Output

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

Sample Input

Sample Output

Source

BestCoder Round #33

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题意：数字1~n，按某种顺序排列，且满足下列某一个条件：（1）a1~ai递增，ai~an递减（2）a1~ai递减，ai~an递增。问有多少种不同的排列。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.

代码：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 2005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r #define FRE(i,a,b)  for(i = a; i <= b; i++) #define FRL(i,a,b)  for(i = a; i < b; i++) #define mem(t, v)   memset ((t) , v, sizeof(t)) #define sf(n)    scanf("%d", &n) #define sff(a,b) scanf("%d %d", &a, &b) #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) #define pf    printf #define DBG   pf("Hi/n") typedef __int64 ll; using namespace std; ll n,p; ll quickmul(ll x,ll m)  //快速乘法，与快速幂相似 {  ll re=0;  while(m){   if(m&1){    re=(re+x)%p;   }   x=(x+x)%p;   m>>=1;  }  return re; } ll pow_m(ll a,ll n) //快速幂 {  ll ret=1;  ll temp=a%p;  while (n){   if (n&1) ret=quickmul(ret,temp)%p;   temp=quickmul(temp,temp)%p;   n>>=1;  }  return ret; } int main() {  int i,j;  while (~scanf("%I64d%I64d",&n,&p))  {   if (n==1)    //特判   {    if (p==1) pf("0/n");    else pf("1/n");    continue;   }   pf("%I64d/n",(pow_m(2,n)-2+p)%p);  }  return 0; }