C. Not Wool Sequences
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
A sequence of non-negative integers a1,?a2,?...,?an of length n is called a wool sequence if and only if there exists two integers l and r(1?≤?l?≤?r?≤?n) such that . In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.
The expression means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".
In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m?-?1 that are not a wool sequence. You should print this number modulo 1000000009 (109?+?9).
Input
The only line of input contains two space-separated integers n and m (1?≤?n,?m?≤?105).
Output
Print the required number of sequences modulo 1000000009 (109?+?9) on the only line of output.
Examples
input
Copy
3 2
output
Copy
6
Note
Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3).
题意：
从０～２＾ｍ－１个数选ｎ个数排列，保证找不到任意区间的异或和等于０，问满足的排列？
分析；
我们对ａ［ｉ］求前缀异或和ｓ［ｉ］，那么ａ［ｉ］＾……＾ａ［ｊ］＝ｓ［ｊ］＾ｓ［ｉ－１］！＝０，所以我们只需保证ｓ［ｊ］和ｓ［ｉ－１］不相同即可。我们还知道如果把任意一段前缀异或和统计起来，然后去重，他肯定包含０～２＾ｍ－１个数。
我们知道０不能选，所有从（２＾ｍ－１）个数选择ｎ个不同的即可。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long long ll;
///b&1取出b二进制位下表示的最低位
///b>>1舍弃最低位
const ll MOD=1e9+9;
ll quick_qow(ll a,ll b,ll p)///计算a^b %p
{ll ans=1%p;///ans不能为1，b为0，p为1的情况不成立for(;b;b>>=1){if(b&1) ans=(ans*a)%p ;a=a*a%p ;}return ans;
}
int main()
{LL n,m;scanf("%lld%lld",&n,&m);LL k=quick_qow(2,m,MOD);LL ans=1;for(int i=0;i<n;i++){ans=(ans*(k-1-i))%MOD;}printf("%lld\n",ans);return 0;
}