首页 > 代码库 > HDU-4927 Series 1

HDU-4927 Series 1

2024-07-16 08:33:04   220人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=4927

同学用java写的大整数相减

Series 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 295    Accepted Submission(s): 100

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
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);        }    }}

 

 


联系
我们