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Series 1

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

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

Author

BUPT

系数是杨辉三角并且是高精度，用java处理。

//515MS	9976K
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner cin=new Scanner(System.in);
		int t;
		t=cin.nextInt();
		for(int cas=1;cas<=t;cas++){
			BigInteger ans=BigInteger.valueOf(0);
			int n;
			n=cin.nextInt();
			BigInteger yanghui[]=new BigInteger[n];
			BigInteger s[]=new BigInteger[n+2];
			for(int i=0;i<n;i++)
				s[i]=cin.nextBigInteger();
			if (n == 1) {
                System.out.println(s[0]);
                continue;
            }
			yanghui[0]=BigInteger.valueOf(1);
			for(int i=1;i<n;i++){
				BigInteger x1=BigInteger.valueOf(i);
				BigInteger x2=BigInteger.valueOf(n-i);
				yanghui[i]=yanghui[i-1].multiply(x2).divide(x1);
			}
			boolean flag=true;
			for(int i=n-1;i>=0;i--){
				if(flag){ans=ans.add(yanghui[i].multiply(s[i]));flag=false;}
				else {ans=ans.subtract(yanghui[i].multiply(s[i]));flag=true;}
			}
			System.out.println(ans);
		}
	}
}

HDU 4927 Series 1 杨辉三角形