题目链接：

Series 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 423    Accepted Submission(s): 146

Problem Description

Let A be an integral series {A₁, A₂, . . . , A_n}.

The zero-order series of A is A itself.

The first-order series of A is {B₁, B₂, . . . , B_n-1},where B_i = A_i+1 - A_i.

The ith-order series of A is 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 that integer.

Input

The 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, . . . , A_n. (0<=A_i<=10⁵)

Output

For each test case, output the required integer in a line.

Sample Input

2
3
1 2 3
4
1 5 7 2

Sample Output

0
-5

Source

2014 Multi-University
Training Contest 6

题目分析：

公式：C(n-1,0)*a[n]-C(n-1,1)*a[n-1]+C(n-1,2)*a[n-2]+...+C(n-1,n-1)*a[n]。

用C(n,k)=C(n,k-1)*(n-k+1)/k即可快速得到一行的二项式系数。

第一道Java题！

代码如下：

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

public class Main {
    public static void main(String[] args) {
        Scanner cin =  new Scanner(System.in);
        BigInteger[] a;
        a = new BigInteger[3017];
        int t, n;
         t = cin.nextInt();

         while(t--){
            n = cin.nextInt();

            for (int i = 0; i < n; i++)
                a[i] = cin.nextBigInteger();

            BigInteger ans = BigInteger.ZERO;
            BigInteger c =  BigInteger.ONE; 

            for (int i = 0; i < n; i++) {
                BigInteger tmp = c.multiply(a[n-i-1]);

                if (i%2 == 0)
                    ans = ans.add(tmp);
                else
                    ans = ans.subtract(tmp);

                tmp = c.multiply(BigInteger.valueOf(n-i-1));
                c = tmp.divide(BigInteger.valueOf(i+1));
            }

            System.out.println(ans);
        }
    }
}

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