[BZOJ2844]albus就是要第一个出场

嘟嘟嘟(洛谷)


这道题算是利用了线性基的一个性质:对于大小为\(k\)的线性基中亦或出来的一个数\(x\),那么用原数组的数亦或出来的所有数中,\(x\)出现了\(2 ^ {n - k}\)次。
那么就可以算出线性基中比\(x\)小的数有多少个(记为\(cnt\)),然后答案就是\(cnt * 2 ^ {n - k} + 1\)
(其实我现在还不是特别懂)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int maxN = 31;
const ll mod = 10086;
inline ll read()
{
	ll ans = 0;
	char ch = getchar(), last = ' ';
	while(!isdigit(ch)) {last = ch; ch = getchar();}
	while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
	if(last == '-') ans = -ans;
	return ans;
}
inline void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

int n;
int p[maxN];
In void insert(int x)
{
	for(int i = maxN; i >= 0; --i)	if((x >> i) & 1)
	{
		if(p[i]) x ^= p[i];
		else {p[i] = x; return;}
	}
}
int b[maxN], cnt = 0;

In ll quickpow(ll a, ll b)
{
	ll ret = 1;
	for(; b; b >>= 1, a = a * a % mod)
		if(b & 1) ret = ret * a % mod;
	return ret;
}

int main()
{
	n = read(); ll x;
	for(int i = 1; i <= n; ++i) x = read(), insert(x);
	for(int i = 0; i <= maxN; ++i) if(p[i]) b[cnt++] = i;
	x = read();
	ll ans = 0;
	for(int i = 0; i < cnt; ++i) if((x >> b[i]) & 1) ans += (1 << i);
	ans %= mod;	
	write((ans * quickpow(2, n - cnt) + 1) % mod), enter;
	return 0;
}
posted @ 2019-01-11 15:27  mrclr  阅读(91)  评论(0编辑  收藏  举报