題目傳送門

題目大意

給出 \(n,k\) 以及 \(a_{1,2,...,n}\) ，求有多少個 \(m_{1,2,...,n}\) 滿足 \(\forall i,m_i\le a_i\) 且 \(\oplus_{i=1}^{n} m_i=k\) 。

\(n\le 50,a_i\le 2^{31}-1\)

思路

這個題目真的很神仙。。。

首先你要想到一點，就是對於二進位制下的數，肯定是前面一段都相同，突然某一位 \(a_i=1\) 你 \(m_i=0\) 那麼 \(m_i\) 你後面就可以亂選了。然後根據這個我們可以設狀態 \(dp[i][len][pre]\) 表示到第 \(i\) 個數，你前面 \(len\)

\(\texttt{Code}\)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 1000000003

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}

int n,m,a[55],pw[35],dp[55][35][2];

int dfs (int i,int pre,int len){
	pre &= (~((1 << len) - 1));
	if (i > n) return !pre;
	int k = (pre & (1 << len)) ? 1 : 0,res = 0,now = 0;
	if (~dp[i][len][k]) return dp[i][len][k];
	for (Int j = 31;~j;-- j)
		if (a[i] & (1 << j)){
			res = add (res,mul (pw[min (len,j)],dfs (i + 1,pre ^ now,max (len,j))));
			now |= (1 << j);	
		}
	return dp[i][len][k] = res;
}

signed main(){
	pw[0] = 1;
	for (Int i = 1;i <= 31;++ i) pw[i] = (pw[i - 1] << 1) % mod;
	while (~scanf ("%d%d",&n,&m) && (n || m)){
		memset (dp,-1,sizeof (dp));
		for (Int i = 1;i <= n;++ i) read (a[i]),a[i] ++;
		write (dfs (1,m,0)),putchar ('\n');
	}
 	return 0;
}