HDU 4927 Series 1(瞎搞)

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最新推荐文章于 2017-08-30 19:50:40 发布

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分类专栏： HDU刷题数学----瞎搞 2014多校联合训练

本文链接：https://blog.youkuaiyun.com/fulongxu/article/details/38425435

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本文介绍了一个关于序列差分计算的问题，给出了使用Java实现的具体代码。该算法通过计算序列的高阶差分找到(n-1)阶差分的值，利用了杨辉三角形的特性并采用BigInteger类来处理可能的大整数运算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

不化简直接进行计算，比如：1，2，3。一遍之后：（2-1），（3-2）。然后：（3-2）-（2-1）。化简的3-2*2+1.

以此写几组就会发现，系数满足杨辉三角、、但是需要java高精度。要便算边求组合数，否则会超时啊、、sad

Series 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 456 Accepted Submission(s): 160

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

Sample Output

  
   0
-5

Source

2014 Multi-University Training Contest 6

import java.util.Scanner;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigInteger[] bit = new BigInteger[3300];
        BigInteger ans,temp,sn;
    
        int i,cas,n,j,k;
        cas = cin.nextInt();
        for(i = 1;i <= cas;i ++)
        {
            ans = BigInteger.ZERO;
            temp = BigInteger.ONE;
            n = cin.nextInt();
            sn = BigInteger.valueOf(n-1);
            for(j = 1;j <= n;j ++)
                bit[j] = cin.nextBigInteger();
            for(j = 0;j < n;j ++)
            {
                if(j%2 == 0)
                    ans = ans.add(temp.multiply(bit[n-j]));
                else 
                    ans = ans.subtract(temp.multiply(bit[n-j]));
                temp = temp.multiply(sn).divide(BigInteger.valueOf(j+1));
                sn = sn.subtract(BigInteger.ONE);
            }
            System.out.println(ans);
        }
    }
}