Blog of RuSun

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

P4132 [BJOI2012]算不出的等式

P4132 [BJOI2012]算不出的等式

相当于板子题了。

查看代码
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#include <cstdio>
#include <algorithm>
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);
}
typedef unsigned int UI;
typedef unsigned long long ULL;
struct Node
{
ULL f, u, r;
Node (ULL _f = 0, ULL _u = 0, ULL _r = 0) : f(_f), u(_u), r(_r) {}
friend Node operator * (Node a, Node b)
{ return Node(a.f + b.f + a.u * b.r, a.u + b.u, a.r + b.r); }
};
Node binpow (Node b, UI k)
{
Node res = Node();
for (; k; k >>= 1, b = b * b)
if (k & 1) res = res * b;
return res;
}
Node ugcd(UI a, UI b, UI c, UI n, Node U, Node R)
{
b %= c;
if (a >= c) return ugcd(a % c, b, c, n, U, binpow(U, a / c) * R);
UI m = ((ULL)a * n + b) / c;
if (m == 0) return binpow(R, n);
swap(U, R);
return binpow(U, (c - b - 1) / a) * R * ugcd(c, c - b - 1, a, m - 1, U, R) * binpow(U, n - ((ULL)c * m - b - 1) / a);
}
int main ()
{
UI p, q;
read(p, q);
Node U = Node(0, 1, 0), R = Node(0, 0, 1);
write(ugcd(q, 0, p, p >> 1, U, R).f + ugcd(p, 0, q, q >> 1, U, R).f);
return 0;
}

打表发现规律有

$$
ANS = \begin {cases}
\frac {pq} 4 & p = q \\
\frac {(p - 1) (q - 1)} 4 & p \neq q
\end {cases}
$$