solution

$f[i][j][k][num][p]:2^p$ 选择了 $i$ 个，前 $p-1$ 位 $(2^0{2}^{p-1})$ 已经选了$j$个，低位向高位进位上来 $k$，前 $p$ 位已经确定有 $num$ 个位置为 $1$，所有的情况和

$g[j][k][num][p]:$ 前 $p-1$ 位一共选了 $j$ 个数，向 $p$ 位进了 $k$，前 $p-1$ 位一共确定有 $num$ 个位置为 $1$，所有的情况和

考虑彼此转移更新
$$f[i][j][k][num][p]=g[j][k][num-(i+k)\%2][p]\ast C(n-j,i)\ast (v[p]{)}^i$$$$g[i+j][(i+k)/2][num][p+1]+=f[i][j][k][num][p]$$

code

#include <cstdio>
#define maxn 35
#define maxm 105
#define mod 998244353
int N, M, K;
int v[maxm], fac[maxn], inv[maxn];
int Pow[maxm][maxn];
int g[maxn][maxn][maxn][maxm];
int f[maxn][maxn][maxn][maxn][maxm];int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = 1ll * ans * x % mod;x = 1ll * x * x % mod;y >>= 1;}return ans;
}void init() {fac[0] = inv[0] = 1;for( int i = 1;i <= N;i ++ ) fac[i] = 1ll * fac[i - 1] * i % mod;inv[N] = qkpow( fac[N], mod - 2 );for( int i = N - 1;i;i -- ) inv[i] = 1ll * inv[i + 1] * ( i + 1 ) % mod;
}int C( int n, int m ) { return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }int main() {scanf( "%d %d %d", &N, &M, &K );init();for( int i = 0;i <= M;i ++ ) {scanf( "%d", &v[i] );Pow[i][0] = 1;for( int j = 1;j <= N;j ++ ) Pow[i][j] = 1ll * Pow[i][j - 1] * v[i] % mod;}g[0][0][0][0] = 1;for( int p = 0;p <= M;p ++ )for( int i = 0;i <= N;i ++ )for( int j = 0;j <= N - i;j ++ )for( int k = 0;k <= N;k ++ )for( int num = 0;num <= K;num ++ ) {if( num - ( i + k ) % 2 < 0 ) continue;f[i][j][k][num][p] = 1ll * g[j][k][num - (i + k) % 2][p] * Pow[p][i] % mod * C( N - j, i ) % mod;g[i + j][(i + k) / 2][num][p + 1] = ( g[i + j][(i + k) / 2][num][p + 1] + f[i][j][k][num][p] ) % mod;}int ans = 0;for( int i = 0;i <= N;i ++ ) for( int k = 0;k <= N;k ++ )for( int num = 0;num <= K;num ++ )if( num + __builtin_popcount( ( i + k ) / 2 ) > K ) break;else ans = ( 1ll * ans + f[i][N - i][k][num][M] ) % mod;printf( "%d\n", ans );return 0;
}