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http://acm.hdu.edu.cn/showproblem.php?pid=5187
n
zhx thinks the
ith
i
zhx defines a sequence
{ai}
i
1:
a1..ai
2:
ai..an
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
1000
EOF
For each case, there are two integers
n
p
1≤n,p≤1018
2 233 3 5
2 1HintIn 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
/**
hdu 5187 高速幂高速乘法
题目大意：（转）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。
解题思路：首先是所有递减或所有递增各一种；另外就是满足上列两个条件的情况了。要想满足条件（1）那就仅仅能把最大的n放在i位置，
共同拥有C（1，n-1）+C（2。n-1）+。。。+C（n-2，n-1）即2^(n-1)-2；条件（2）与（1）同样，所以共同拥有(2^(n-1)-2)*2+2=2^n-2.**/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;typedef long long LL;LL n,p;LL qui_mul(LL x,LL m)///高速乘法{ LL re=0; while(m) { if(m&1) { re=(re+x)%p; } x=(x+x)%p; m>>=1; } return re;}LL qui_pow(LL a,LL n)///高速幂{ LL ret=1; LL tem=a%p; while(n) { if(n%1)ret=qui_mul(ret,temp)%p; temp=qui_mul(temp,temp)%p; n>>=1; } return ret;}int main(){ while(~scanf("%I64d%I64d",&n,&p)) { if(n==1) { if(p==1) printf("0\n"); else printf("1\n"); } printf("%I64d\n",(qui_mul(2,n)-2)%p); } return 0;}
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