Blog of RuSun

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

CF383E Vowels

LuoGu: CF383E Vowels

CF: E. Vowels

对于每一个元音字母集合 $a$ ,每一个单词 $b$ ,如果 a & b > 0 则为合法,那么 a & b = 0 则为不合法。考虑不合法的情况,对于一个元音字母集合 $a$ ,其取反后含有子集 $b$ 则 $b$ 贡献,可以用高维前缀和做。

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 << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1 << 24;
int n, f[N];
int main()
{
read(n);
for (int i = 1; i <= n; i++)
{
int x = 0;
char s[5];
scanf("%s", s);
for (int j = 0; j < 3; j++)
s[j] <= 'x' && (x |= 1 << s[j] - 'a');
f[x]++;
}
for (int i = 0; i < 24; i++)
for (int j = 0; j < 1 << 24; j++)
j >> i & 1 && (f[j] += f[j ^ 1 << i]);
int res = 0;
for (int i = 0; i < 1 << 24; i++)
res ^= (n - f[i]) * (n - f[i]);
write(res);
return 0;
}