E. Data Structures Fan
这道题可以先对数组\(a\)进行异或前缀和操作，得到数组\(b\)，并在初始情况下记录下所有\(s[i]\)为0/1的异或和为\(x0\)，\(x1\)，之后处理询问操作。
- 当\(ty=1\)时，需要对\(s[i]\)(\(l≤i≤r\))进行反转，即让\(x0\)和\(x1\)对\(a[i]\)(\(l≤i≤r\))进行异或操作，即
\[x0= x0\oplus(b[r] \oplus b[l - 1]);
\]
\[x1= x1\oplus(b[r] \oplus b[l - 1]);
\]
- 当\(ty=2\)时，输出\(x0\)或\(x1\)。
代码
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PII;
const ll N = 1e5 + 10;
ll t;
ll n, a[N], q;
ll b[N];
string s;
signed main()
{
std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> t;
while(t--)
{
cin >> n;
ll x0 = 0, x1 = 0;
for(ll i = 1;i <= n;i++)
{
cin >> a[i];
b[i] = b[i - 1] ^ a[i];
}
cin >> s;
s = ' ' + s;
for(ll i = 1;i <= n;i++)
{
if(s[i] == '0')
{
x0 ^= a[i];
}
else
{
x1 ^= a[i];
}
}
cin >> q;
while(q--)
{
ll ty;
cin >> ty;
if(ty == 1)
{
ll l, r;
cin >> l >> r;
x0 ^= (b[r] ^ b[l - 1]);
x1 ^= (b[r] ^ b[l - 1]);
}
else
{
ll g;
cin >> g;
if(g == 0)
{
cout << x0 << " ";
}
else
{
cout << x1 << " ";
}
}
}
cout << endl;
}
return 0;
}