poj 2094 多项式求和。

/**
给出多项式 p(x) = an*x^n + an-1*x^(n-1)..... + a1*x + a0;
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

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。
**/

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

        public static int cal(String str, int m) {
               // System.out.println(str+"------->");
               int len = Math.min(str.length(), m);
               int ans = 0, tmp;
               for (int i = 0; i < len; i++) {
                     tmp = str.charAt(i) - '0';
                     ans += tmp * tmp;
              }
               return ans;
       }

        public static void main(String[] args) {
              Scanner cin = new Scanner(System. in);
               int n = cin.nextInt();
              BigInteger l = cin.nextBigInteger();
               int k = cin.nextInt();
               int m = cin.nextInt();
              BigInteger mod = BigInteger. TEN.pow(m);
              BigInteger[] a = new BigInteger[15];
               for (int i = 0; i <= n; i++)
                     a[i] = cin.nextBigInteger();
              BigInteger h[][] = new BigInteger[15][15];
               for (int i = 0; i < Math.min (n + 1, k); i++) {
                     BigInteger res = a[0];
                      for (int j = 1; j <= n; j++) {
                           res = res.multiply(l);
                           res = res.add(a[j]);
                     }
                     res = res.mod(mod);
                     l = l.add(BigInteger. ONE);
                     h[0][i] = res;
                     System. out.println(cal(res.toString(), m));
              }

               if (k > n + 1) {
                      for (int i = 1; i <= n; i++) {
                            for (int j = 0; j <= n - i; j++)
                                  h[i][j] = h[i - 1][j + 1].subtract(h[i - 1][j]);
                     }
              }

              BigInteger pre[] = new BigInteger[15];
               for (int i = 0; i <= n; i++) {
                     pre[i] = h[i][n - i];
              }
               for (int i = n + 1; i < k; i++) {
                     BigInteger res = h[n][0];
                      for (int j = n - 1; j >= 0; j--) {
                           res = pre[j].add(res);
                           res = res.mod(mod);
                           pre[j] = res;
                     }
                     System. out.println(cal(res.toString(), m));
              }
       }

}