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

BZOJ 3771. Triple

#3771. Triple

记 $A, B, C$ ,分别表示一个数重复选择 $1, 2, 3$ 次的方案数。考虑容斥,那么选择一个数的方案为 $A$ ,选择两个数的方案为 $\frac {A * A - B} 2$ ,选择三个数的方案为 $\frac {A * A * A - 3 A * B + 2C} 6$ 。

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 long long LL;
const int N = 1e6 + 10, mod = 998244353, inv2 = 499122177, inv6 = 166374059;
int n, m, rev[N], inv[N], fact[N], ifact[N], sn[N], pm[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void ntt (int *x, int bit, int op)
{
int tot = 1 << bit;
for (int i = 1; i < tot; ++i)
if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
swap(x[rev[i]], x[i]);
for (int mid = 1; mid < tot; mid <<= 1)
{
int w1 = binpow(3, (mod - 1) / (mid << 1));
if (!~op) w1 = binpow(w1);
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
{
int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
}
}
if (~op) return;
int itot = binpow(tot);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * itot % mod;
}
int main ()
{
static int n = 12e4, m, A[N], B[N], C[N];
read(m);
for (int a; m; --m)
{
read(a);
++A[a], ++B[a * 2], ++C[a * 3];
}
int bit = 0;
while (1 << bit < n + n + n - 2) ++bit;
int tot = 1 << bit;
ntt(A, bit, 1), ntt(B, bit, 1), ntt(C, bit, 1);
for (int i = 0; i < tot; ++i)
A[i] = (((LL)A[i] * A[i] % mod * A[i] - 3ll * A[i] * B[i] + 2 * C[i]) % mod * inv6 + ((LL)A[i] * A[i] - B[i]) % mod * inv2 + A[i]) % mod;
ntt(A, bit, -1);
for (int i = 1; i <= n; ++i)
if ((A[i] += mod) %= mod)
write(i), putchar(' '), write(A[i]), puts("");
return 0;
}