poj 2094 多项式求和。

 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 }

poj 2094 多项式求和。