題目

分析

考慮 \(kruskal\) 的過程
我們選邊從高位開始
當前位為 \(0\) 的放一邊，為 \(1\) 的放另一邊
將 \(0\) 的建一棵字典樹， \(1\) 的匹配
因為是異或，那就走相同值的位，算能匹配到的最小值的個數
和與方案數都可以在這裡計算

\(Code\)

#include<cstdio>
using namespace std;
typedef long long LL;

const LL P = 1e9 + 7;
const int N = 100005;
int n , cnt , su , a[N] , c[N] , d[N] , ts[N * 30] , t[N * 30][2];
LL ans = 1;

LL fpow(LL x , LL y)
{
	LL res = 1;
	while (y)
	{
		if (y & 1) res = res * x % P;
		y >>= 1 , x = x * x % P;
	}
	return res;
}

void insert(int x)
{
	int u = 0 , ch;
	for(register int i = 30; i >= 0; i--)
	{
		ch = (x >> i) & 1;
		if (!t[u][ch]) t[u][ch] = ++cnt;
		u = t[u][ch] , ts[u]++;
	}
}

int find(int x)
{
	int u = 0 , ch , res = 0;
	for(register int i = 30; i >= 0; i--)
	{
		ch = (x >> i) & 1;
		if (t[u][ch]) u = t[u][ch];
		else u = t[u][ch ^ 1] , res = res + (1 << i);
	} 
	su = ts[u];
	return res;
}

LL solve(int l , int r , int w)
{
	if (l >= r) return 0;
	if (w == -1) 
	{
		if (r - l - 1 > 0) ans = ans * fpow(r - l + 1 , r - l - 1) % P;
		return 0;
	}
	int tl = 0 , tr = 0;
	for(register int i = l; i <= r; i++) 
	if (a[i] & (1 << w)) d[++tr] = a[i];
	else c[++tl] = a[i];
	for(register int i = 1; i <= tl; i++) a[l + i - 1] = c[i];
	for(register int i = 1; i <= tr; i++) a[l + tl - 1 + i] = d[i];
	int tmp;
	if (!tl || !tr) tmp = 0;
	else
	{
		int num = 0 , f; tmp = 2147483647 , cnt = 0;
		for(register int i = 1; i <= tl; i++) insert(c[i]);
		for(register int i = 1; i <= tr; i++)
		{
			su = 0 , f = find(d[i]);
			if (f < tmp) tmp = f , num = su;
			else if (tmp == f) num += su;
		}
		ans = ans * num % P;
		for(register int i = 0; i <= cnt; i++) ts[i] = t[i][0] = t[i][1] = 0;
	}
	return 1LL * tmp + solve(l , l + tl - 1 , w - 1) + solve(l + tl , r , w - 1);
} 

int main()
{
	freopen("jst.in" , "r" , stdin);
	freopen("jst.out" , "w" , stdout);
	scanf("%d" , &n);
	for(register int i = 1; i <= n; i++) scanf("%d" , &a[i]);
	printf("%lld\n" , solve(1 , n , 30));
	printf("%lld" , ans);
}