HDU4927：Series 1（JAVA大数）

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.

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)

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

2
3
1 2 3
4
1 5 7 2

0
-5

开始没考虑太多，直接暴力一交，WA
暴力最多超时，怎么WA呢，第一个想法就是，大数
大数怎么能忘记JAVA呢？C++写大数太麻烦了，果断JAVA变成水题

import java.math.BigInteger;
import java.util.*;
import java.io.*;

public class Main {

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int a[] = new int[3005];
        for (int cas = 1; cas <= t; cas++) {
            int n = in.nextInt();
            int  i, j;
            for (i = 1; i <= n; ++i) {
                a[i] = in.nextInt();
            }
            BigInteger ans = BigInteger.valueOf(a[n]);
            BigInteger x = BigInteger.valueOf(1);
            BigInteger flag = BigInteger.valueOf(-1);
            n = n -1;
            for(i=1,j=n; i<=n; i++,j--)
            {
                x = x.multiply(BigInteger.valueOf(j)).divide(BigInteger.valueOf(i));
                x = x.multiply(flag);
                ans = ans.add(x.multiply(BigInteger.valueOf(a[j])));
            }
            System.out.println(ans);
        }
    }
}