HDU 4927 Series 1 ( 组合+高精度)

pid=4927">Series 1

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 {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

 

 

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);
			}
 
		}
	}