Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3760 [TJOI2017] 异或和

P3760 [TJOI2017] 异或和

做前缀和,二进制下拆为,注意到一个差值的某一位为 $1$ ,只和这一位以及后面的位有关,具体地,如果这一位两数相反,那么相减不能借位;否则一定要借位。那么即二维偏序,树状数组即可。

查看代码
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#include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
const int N = 1e5 + 10, M = 1e6 + 10, K = 1e6;
bool tr[2][M];
int n, w[N];
void add (bool *tr, int x)
{
for (++x; x <= K; x += x & -x)
tr[x] ^= 1;
}
bool query (bool *tr, int x)
{
bool res = 0;
for (++x; x; x -= x & -x)
res ^= tr[x];
return res;
}
int main ()
{
read(n);
for (int i = 1; i <= n; ++i)
read(w[i]), w[i] += w[i - 1];
int res = 0;
add(tr[0], 0);
for (int k = 0; k < 20; ++k)
{
bool s = 0;
for (int i = 1; i <= n; ++i)
{
bool v = w[i] >> k & 1;
int x = w[i] & (1 << k) - 1;
s ^= query(tr[v ^ 1], x);
s ^= query(tr[v], (1 << k) - 1) ^ query(tr[v], x);
add(tr[v], x);
}
for (int i = 1; i <= n; ++i)
{
bool v = w[i] >> k & 1;
int x = w[i] & (1 << k) - 1;
add(tr[v], x);
}
res |= s << k;
}
write(res);
return 0;
}