大意：
题意不好翻译，英文看懂也不是非常麻烦，就不翻译了。
Problem Description
Let A be an integral series {A 1, A 2, . . . , A n}. The zero-order series of A is A itself. The first-order series of A is {B 1, B 2, . . . , 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 A 1, A 2, . . . , A n. (0<=A i<=10 5)
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
思路：
比赛中楠姐非常快就推出来公式了，想把杨辉三角预处理出来，然后发现BigInteger大小爆内存了。。
。
。非常无语
然后又想在暴力的基础上去优化。然后一直T到死。
。。 比赛结束也没搞出来。
赛后才知道。杨辉三角是能够直接用组合公式推出来的。。。
杨辉三角的第n行的第m个数为组合数c[n-1][m-1]。
应用c[n][m] = c[n][m-1]*(n-m+1)/m，就能够高速递推出第n行的数，这样既避免了打表会出现的爆内存。也省去了暴力好多的时间。。。。
。
还是太年轻 哎。。。
import java.io.*; import java.math.*; import java.util.*; public class Main { static BigInteger coe[][] = new BigInteger [3010][3010]; public static void main(String[] args) throws IOException{ Scanner cin = new Scanner(System.in); BigInteger []a = new BigInteger[3010]; BigInteger []c = new BigInteger[3010]; int T; T = cin.nextInt(); while(T-- > 0){ int n; n = cin.nextInt(); for(int i = 1; i <= n; ++i){ a[i] = cin.nextBigInteger(); } BigInteger ans = BigInteger.ZERO; c[0] = BigInteger.ONE; ans = ans.add(c[0].multiply(a[n])); int t = -1; for(int i = 1; i < n; ++i){ BigInteger t1 = BigInteger.valueOf(n).subtract(BigInteger.valueOf(i)); BigInteger t2 = BigInteger.valueOf(i); c[i] = c[i-1].multiply(t1).divide(t2); ans = ans.add(c[i].multiply(a[n-i]).multiply(BigInteger.valueOf(t))); t *= -1; } System.out.println(ans); } } }