2014ACM集训13级PK赛5-Easy Task

Description

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

06 230 0 1
 

 

求导的题目，还算轻松

#include <string.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int main()
{
int N;
scanf ("%d",&N);
while (N--)
{
int n,i;
scanf ("%d",&n);

for (i = n;i >= 0;i--)
{
int tmp;
scanf ("%d",&tmp);

if (i > 0)
printf ("%d",tmp * i);
else if (n == 0)
printf ("0");
if (i > 1)
printf (" ");
else if (i == 0)
printf ("\n");
}

}
return 0;
}