HDU 4927 Series 1

Problem Descriptionlet A is an integral series {A1, A2, ..., an}.

The Zero-order series of A is a itself.

The First-order series of A is {B1, B2, ..., bn-1},where Bi = Ai+1-ai.

The Ith-order series of A are the First-order series of its (i-1) Th-order series (2<=i<=n-1).

Obviously, the (n-1) Th-order series of A is a single integer. Given A, figure out of that integer.
Inputthe input consists of several test cases. The first line of input gives the number of test cases T (t<=10).

For each test case:
The first line contains a single integer n (1<=n<=3000), which denotes the length of series A.
The second line consists of n integers, describing A1, A2, ..., an. (0<=ai<=105)

Outputfor each test case, output the required integer in a line.
Sample Input

231 2 341 5 7 2

Sample Output

0-5

Test instructions: What is the number of final merges?

Idea: Java high-precision, pushed out after the discovery is the coefficient is Yang Hui triangle. Calculation results after processing the coefficients

Import Java.math.biginteger;import java.util.*;import java.io.*;/** * Created by Acer on 14-8-7.        */public class Main {public static void Main (string[] args) {Scanner cin = new Scanner (system.in);        int t, N;        t = Cin.nextint ();            while (t! = 0) {t--;            n = cin.nextint ();            BigInteger arr[] = new Biginteger[n];            for (int i = 0; i < n; i++) {Arr[i] = Cin.nextbiginteger ();                } if (n = = 1) {System.out.println (arr[0]);            Continue            } BigInteger ans = new BigInteger ("0");            BigInteger c[] = new Biginteger[n + 2];            BigInteger T1 = new BigInteger ("0");            BigInteger t2 = new BigInteger ("0");            C[0] = biginteger.valueof (1);                for (int i = 1; i < n; i++) {t1 = biginteger.valueof (n-i);                T2 = biginteger.valueof (i); C[i] = c[i-1].multiply (T1). Divide (T2);            } int flag = 1; for (int i = n-1; I >= 0; i--) {if (flag = =-1) ans = ans.subtract (arr[i].multiply                (C[i]));                else ans = ans.add (arr[i].multiply (C[i]));            Flag *=-1;        } System.out.println (ANS); }    }}

HDU 4927 Series 1