Blog of RuSun

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

P7961 [NOIP2021] 数列

P7961 [NOIP2021] 数列

从小到大考虑每个 $v$ 选择几个,注意到二进制表示是不好维护,但是从小到大枚举更小的位置是确定的,因此用 DP 两维状态表示当前位置一下有多少个 $1$ ,对其他部分影响直接存在状态交给后面处理。

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);
}
typedef long long LL;
const int N = 40, M = 110, mod = 998244353;
int n, m, h, cnt[N], C[N][N], p[M][N], f[M][N][N][N >> 1];
void add (int &x, LL k) { x = (x + k) % mod; }
void add (int &x, int k) { x += k - mod, x += x >> 31 & mod; }
void init ()
{
for (int i = 0; i <= n; ++i) C[i][0] = 1;
for (int i = 1; i < N; ++i)
for (int j = 1; j <= i; ++j)
add(C[i][j] = C[i - 1][j], C[i - 1][j - 1]);
for (int i = 1; i < N; ++i)
cnt[i] = cnt[i >> 1] + (i & 1);
for (int i = 0, a; i <= m; ++i)
{
read(a);
p[i][0] = 1;
for (int j = 1; j <= n; ++j)
p[i][j] = (LL)p[i][j - 1] * a % mod;
}
}
int main ()
{
read(n, m, h);
init();
f[0][0][0][0] = 1;
for (int i = 0; i <= m; ++i) for (int j = 0; j <= n; ++j) for (int k = 0; k <= h; ++k) for (int s = 0; s <= n >> 1; ++s)
for (int t = 0; j + t <= n; ++t) add(f[i + 1][j + t][k + (t + s & 1)][t + s >> 1], (LL)p[i][t] * C[n - j][t] % mod * f[i][j][k][s]);
int res = 0;
for (int k = 0; k <= h; ++k) for (int s = 0; s <= n >> 1; ++s)
if (k + cnt[s] <= h) add(res, f[m + 1][n][k][s]);
write(res);
return 0;
}