HDU5187 zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2388    Accepted Submission(s): 755

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

 

 

Recommend

hujie

 

贪心？

答案是2^n-2

只是用来练快速乘的

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#define LL long long
using namespace std;
const int mxn=100010;
LL n,p;
LL ksmul(LL a,LL b){
    a%=p;b%=p;LL res=0;
    while(b){
        if(b&1){res+=a;if(res>p)res-=p;}
        a=(a<<1)%p;
        b>>=1;
    }
    return res;
}
LL ksm(LL x,LL k){
    LL res=1;
    while(k){
        if(k&1)res=ksmul(res,x);
        x=ksmul(x,x);
        k>>=1;
    }
    return res;
}
int main(){
    while(scanf("%lld%lld",&n,&p)!=EOF){
        if(n==1){
            if(p==1)printf("0\n");
            else printf("1\n");
            continue;
        }
        LL ans=ksm(2,n);ans-=2;
        if(ans<0)ans=((ans%p)+p)%p;
        printf("%lld\n",ans);
    }
    return 0;
}