ZOJ2969 Easy Task java

題目鏈接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2969

                                           Easy Task

Calculating the derivation of a polynomial is an easy task. Given a function f(x) , we use (f(x))' to denote its derivation. We use x^n to denote xn. To calculate the derivation of a polynomial, you should know 3 rules: 
(1) (C)'=0 where C is a constant. 
(2) (Cx^n)'=C*n*x^(n-1) where n>=1 and C is a constant. 
(3) (f1(x)+f2(x))'=(f1(x))'+(f2(x))'. 
It is easy to prove that the derivation a polynomial is also a polynomial.

Here comes the problem, given a polynomial f(x) with non-negative coefficients, can you write a program to calculate the derivation of it?

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 1000) which is the number of test cases. And it will be followed by T consecutive test cases.

There are exactly 2 lines in each test case. The first line of each test case is a single line containing an integer N (0 <= N <= 100). The second line contains N + 1 non-negative integers, CN, CN-1, ..., C1, C0, ( 0 <= Ci <= 1000), which are the coefficients of f(x). Ci is the coefficient of the term with degree i in f(x). (CN!=0)

Output

For each test case calculate the result polynomial g(x) also in a single line. 
(1) If g(x) = 0 just output integer 0.otherwise 
(2) suppose g(x)= Cmx^m+Cm-1x^(m-1)+...+C0 (Cm!=0),then output the integers Cm,Cm-1,...C0.
(3) There is a single space between two integers but no spaces after the last integer. 

Sample Input

3

0

10

2

3   2   1

3

10   0   1   2

Sample Output

0

6   2

30   0   1

題意：

求多項式的導數的係數。每個測試用例中都有2行。每個測試用例的第一行是包含整數 N (0 = N = 100) 的單行。第二行包含 N + 1 非負整數, CN, CN-1,.., C1, C0, 這是 f (x) 的係數。如題目例子，N爲2，第二行爲3 2 1，可表示爲數學表達式3x^2+2x^1+1x^0=6^x+2+0;

所以輸出6 2.

思路：

導數的定義。

代碼：

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		while(sc.hasNext()){
			int t=sc.nextInt();
			for(int i=0;i<t;i++){
				int n=sc.nextInt();
				int[] a=new int[n+1];
				int k=n;
				for(int j=n;j>=0;j--){
					a[j]=sc.nextInt();
				}
				if(k==0){
					System.out.print(0);
				}
				while(k!=0){
					System.out.print(a[k]*k);
					if(k!=1){
						System.out.print(" ");
					}
					k--;
				}
				System.out.println();
			}
		}
	}
}