ZOJ Problem Set – 2969 Easy Task

Easy Task

Time Limit: 1 Second      Memory Limit: 32768 KB

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 Tconsecutive 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

301023 2 1310 0 1 2

Sample Output

06 230 0 1

Author: CAO, Peng
Source: The 5th Zhejiang Provincial Collegiate Programming Contest

SOURCE CODE:

#include<iostream>
#include<sstream>
using namespace std;

int main()
{
int cases;cin>>cases;
for(int time = 1; time <= cases; time++)
    {
int N;cin>>N;
for(int c = 0, num; cin>>num && c < N; c++)
        {
if(c != 0)
                cout<<" ";
            cout<<num * (N - c);
        }
        cout<<(N == 0 ? "0":"")<<endl;
    }
return 0;
}