可求得通项$F_n=\frac{a^n-1}{a-1}$，一个等比数列的前$n$项和，$value(s)=F_{sum(s)}$。

题目要我们求的是$Answer(K)=\sum _{s\in subsetofSand|s|=K}value(s)$，我们考虑先计算$F_n$中$a^{sum(s)}$部分。

所以我们可以写出生成函数${\prod }_{i=1}^n(1+a^{s_i})$，递归分治进行$FFT$，使这个式子变成一个多项式。

对于$an{s}_k=\frac{[x^k]-C_n^k}{a-1}$，对于每个$k$有$C_n^k$种方案。

然后这题稍卡精度，用$longdouble$就好了。

#include <bits/stdc++.h>using namespace std;#define double long doublestruct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}
};Complex operator + (const Complex &a, const Complex &b) {return Complex(a.r + b.r, a.i + b.i);
}Complex operator - (const Complex &a, const Complex &b) {return Complex(a.r - b.r, a.i - b.i);
}Complex operator * (const Complex &a, const Complex &b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
}Complex operator / (const Complex &a, const Complex &b) {return Complex((a.r * b.r + a.i * b.i) / (b.r * b.r + b.i * b.i), (a.i * b.r - a.r * b.i) / (b.r * b.r + b.i * b.i));
}typedef long long ll;const int N = 1e6 + 10;int r[N];Complex x[N], y[N];void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}void FFT(Complex *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}const double pi = acos(-1.0);for (int mid = 1; mid < lim; mid <<= 1) {Complex wn = Complex(cos(pi / mid), rev * sin(pi / mid));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {Complex w = Complex(1, 0);for (int k = 0; k < mid; k++, w = w * wn) {Complex x = f[cur + k], y = w * f[cur + mid + k];f[cur + k] = x + y, f[cur + mid + k] = x - y;}}}if (rev == -1) {for (int i = 0; i < lim; i++) {f[i].r /= lim;}}
}const int mod = 100003;int n, a, m, s[N];vector<int> f[N];int fac[N], inv[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans;
}ll C(int n, int m) {if(n < 0 || m < 0 || m > n) return 0;if(m == 0 || m == n)    return 1;return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}void init() {fac[0] = 1;for(int i = 1; i < 100003; i++)fac[i] = 1ll * fac[i - 1] * i % mod;inv[100003 - 1] = quick_pow(fac[100003 - 1], mod - 2);for(int i = 100003 - 2; i >= 0; i--)inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}void solve(int rt, int l, int r) {if (l == r) {f[rt].push_back(1);f[rt].push_back(s[l]);return ;}int mid = l + r >> 1;solve(rt << 1, l, mid);solve(rt << 1 | 1, mid + 1, r);int len1 = mid - l + 1, len2 = r - mid;for (int i = 0; i <= len1; i++) {x[i] = Complex(f[rt << 1][i], 0);}for (int i = 0; i <= len2; i++) {y[i] = Complex(f[rt << 1 | 1][i], 0);}int lim = 1;while (lim <= r - l + 1) {lim <<= 1;}get_r(lim);FFT(x, lim, 1);FFT(y, lim, 1);for (int i = 0; i < lim; i++) {x[i] = x[i] * y[i];}FFT(x, lim, -1);for (int i = 0; i <= r - l + 1; i++) {f[rt].push_back(ll(x[i].r + 0.5) % mod);}for (int i = 0; i < lim; i++) {x[i] = y[i] = Complex(0, 0);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d %d", &n, &a, &m);for (int i = 1; i <= n; i++) {scanf("%d", &s[i]);s[i] = quick_pow(a, s[i]);}solve(1, 1, n);init();int inv = quick_pow(a - 1, mod - 2);for (int i = 1, k, ans; i <= m; i++) {scanf("%d", &k);ans = 1ll * (f[1][k] - C(n, k)) * inv % mod;printf("%d\n", (ans % mod + mod) % mod);}return 0;
}