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  • HDU 4927 Series 1

    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

    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);
            }
        }
    }



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  • 原文地址:https://www.cnblogs.com/lcchuguo/p/5376952.html
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