CF1322B Present

传送门


题意：给\(n\)个数，让你求任意两个数之和的异或和。（\(n \leqslant 4\times 10^5, a_i \leqslant 10^7\)）


这题挺有意思的，差点就想出来了。

我们按位考虑，看加和（记为\(sum\)）在这一位上的1是偶数个还是奇数个。

那么对于第\(k\)位，把所有数对\(2^{k+1}\)取模，那么如果\(sum\)的第\(k\)位是1的话，\(sum\)必然属于\([2^k, 2^{k+1})\)或\([2^{k+1}+2^k,2^{k+1}+2^{k+1})\)（即第\(k\)位是否发生了进位）。

所以我们可以在每次取模后排序，然后枚举每一个数，用双指针（或二分）算出和当前数的和在上述区间内的数对个数。

时间复杂度\(O(n\log n\log C)\).

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 4e5 + 5;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
}

int n, a[maxn], b[maxn];

In int calc(int L, int R)		//sum in [L,R)
{
	ll ret = 0;
	for(int i = 1, l = n + 1, r = n; i <= n; ++i)
	{
		while(l > 1 && b[i] + b[l - 1] >= L) l--;
		while(r && b[i] + b[r] >= R) r--;
		if(l <= r) ret += (r - l + 1) - (l <= i && i <= r);	//将数对(i,i)去掉 
	}
	return (ret >> 1) & 1;
}
In int solve(int x)
{
	for(int i = 1; i <= n; ++i) b[i] = a[i] % (x << 1);
	sort(b + 1, b + n + 1);
	return calc(x, x << 1) ^ calc(x + (x << 1), x << 2);
}

int main()
{
//	MYFILE();
	n = read();
	for(int i = 1; i <= n; ++i) a[i] = read();
	int ans = 0;
	for(int i = 0; i < 30; ++i) ans |= (solve(1 << i) << i);
	write(ans), enter;
	return 0;
}