• hdu4927 Series 1(组合+公式 Java大数高精度运算)


    题目链接:


    Series 1

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 423    Accepted Submission(s): 146

    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
     
    Source

    题目分析:
    公式:C(n-1,0)*a[n]-C(n-1,1)*a[n-1]+C(n-1,2)*a[n-2]+...+C(n-1,n-1)*a[n]。
    用C(n,k)=C(n,k-1)*(n-k+1)/k就可以高速得到一行的二项式系数。


    第一道Java题!

    代码例如以下:
    import java.util.Scanner;
    import java.math.BigInteger;
    
    public class Main {
        public static void main(String[] args) { 
            Scanner cin =  new Scanner(System.in); 
            BigInteger[] a;
            a = new BigInteger[3017];
            int t, n;
             t = cin.nextInt();
    
             while(t--){
                n = cin.nextInt();
    
                for (int i = 0; i < n; i++)
                    a[i] = cin.nextBigInteger();
                
                BigInteger ans = BigInteger.ZERO;
                BigInteger c =  BigInteger.ONE; 
    
                for (int i = 0; i < n; i++) {
                    BigInteger tmp = c.multiply(a[n-i-1]);
    
                    if (i%2 == 0)
                        ans = ans.add(tmp);
                    else
                        ans = ans.subtract(tmp);
    
                    tmp = c.multiply(BigInteger.valueOf(n-i-1));
                    c = tmp.divide(BigInteger.valueOf(i+1));
                }
    
                System.out.println(ans);
            }
        } 
    }



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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/6962218.html
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