JZOJ 5352. 【NOIP2017提高A组模拟9.7】计数题

题目

分析

考虑 \(kruskal\) 的过程

我们选边从高位开始

当前位为 \(0\) 的放一边，为 \(1\) 的放另一边

将 \(0\) 的建一棵字典树， \(1\) 的匹配

因为是异或，那就走相同值的位，算能匹配到的最小值的个数

和与方案数都可以在这里计算

\(Code\)

#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);
}