HDU 4927 Series 1

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

题意：求最后合并的数是多少

思路：JAVA高精度，推出来后发现是系数是杨辉三角。处理出系数后计算结果

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);
        }
    }
}