HDU 4927 Series 1 ( 组合+高精度)

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, . . . , An}.

The zero-order series of A is A itself.

The first-order series of A is {B1, B2, . . . , Bn-1},where Bi = Ai+1 - Ai.

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, . . . , An. (0<=Ai<=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
[m] = c
[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
));
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);
}

}
}