Problem Description
As one of the most powerful brushes, zhx is required to give his juniors
problems.zhx thinks the
problem’s difficulty is
. He wants to arrange these problems in a beautiful way.zhx defines a sequence
beautiful if there is an
that matches two rules below:1:
i
are monotone decreasing or monotone increasing.2:
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
.
Input
Multiply test cases(less than ). Seek as the end of the file.For each case, there are two integers
and
separated by a space in a line. (
18
)
Output
For each test case, output a single line indicating the answer.
Sample Input
2 2333 5
Sample Output
21
思路：枚举 减 减 ； 减 加 ; 加 减 加 加 一共 2^(n-1)*2-2
1 2^(n-1) -2 2^(n-1) -2 1
// 2^n-2#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)typedef __int64 ll;#define fre(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 pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1001ll n,mod;ll fdd(ll x,ll m) //计算 x*m 居然不能直接算，，否者会爆__int64 {ll ans=0;while(m){if(m&1) ans=(ans+x)%mod;x=(x+x)%mod;m>>=1;} return ans;}ll pow_(ll n,ll m){ll ans=1;while(m) {if(m&1) ans=fdd(ans,n); //计算 ans*nn=fdd(n,n);//计算 n*nm>>=1; }return (ans-2+mod)%mod;}int main(){while(~scanf("%I64d%I64d",&n,&mod)){ll ans=2;if(n==1){ans=n%mod;pf("%I64d\n",ans);continue;}printf("%I64d\n",pow_(ans,n));}return 0;}
我无所事事的度过了今天，是昨天死去的人们所期望的明天。
