ZOJ Problem Set - 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 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
3 0 10 2 3 2 1 3 10 0 1 2
Sample Output
0 6 2 30 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;
}
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