1 /**
2 给出多项式 p(x) = an*x^n + an-1*x^(n-1)..... + a1*x + a0;
3 给定n，l，k，m 计算 x 从 l到 l+k-1 的p（x）的后m 位的平方的和
4
5 用差分序列 ，先计算出前 n项 构造出差分表。。后边的k-n个 用递推可得，从下往上递推。
6 **/
7
8 import java.math.BigInteger;
9 import java.util.Scanner;
10
11 public class Main {
12
13 public static int cal(String str, int m) {
14 // System.out.println(str+"------->");
15 int len = Math.min(str.length(), m);
16 int ans = 0, tmp;
17 for (int i = 0; i < len; i++) {
18 tmp = str.charAt(i) - '0';
19 ans += tmp * tmp;
20 }
21 return ans;
22 }
23
24 public static void main(String[] args) {
25 Scanner cin = new Scanner(System. in);
26 int n = cin.nextInt();
27 BigInteger l = cin.nextBigInteger();
28 int k = cin.nextInt();
29 int m = cin.nextInt();
30 BigInteger mod = BigInteger. TEN.pow(m);
31 BigInteger[] a = new BigInteger[15];
32 for (int i = 0; i <= n; i++)
33 a[i] = cin.nextBigInteger();
34 BigInteger h[][] = new BigInteger[15][15];
35 for (int i = 0; i < Math.min (n + 1, k); i++) {
36 BigInteger res = a[0];
37 for (int j = 1; j <= n; j++) {
38 res = res.multiply(l);
39 res = res.add(a[j]);
40 }
41 res = res.mod(mod);
42 l = l.add(BigInteger. ONE);
43 h[0][i] = res;
44 System. out.println(cal(res.toString(), m));
45 }
46
47 if (k > n + 1) {
48 for (int i = 1; i <= n; i++) {
49 for (int j = 0; j <= n - i; j++)
50 h[i][j] = h[i - 1][j + 1].subtract(h[i - 1][j]);
51 }
52 }
53
54 BigInteger pre[] = new BigInteger[15];
55 for (int i = 0; i <= n; i++) {
56 pre[i] = h[i][n - i];
57 }
58 for (int i = n + 1; i < k; i++) {
59 BigInteger res = h[n][0];
60 for (int j = n - 1; j >= 0; j--) {
61 res = pre[j].add(res);
62 res = res.mod(mod);
63 pre[j] = res;
64 }
65 System. out.println(cal(res.toString(), m));
66 }
67 }
68
69 }
