BNU16488：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, C_N, C_N-1, ..., C₁, C₀, ( 0 <= C_i <= 1000), which are the coefficients of f(x). C_i is the coefficient of the term with degree i in f(x). (C_N!=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)= C_mx^m+C_m-1x^(m-1)+...+C₀ (C_m!=0),then output the integers C_m,C_m-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

题意：给出n，代表多项式有n+1项，要求多项式求导后每一项的系数，常数项就不用输出

思路：直接水过

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[105],b[105];

int main()
{
    int t,i,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i = 0;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i] = a[i]*(n-i);
        }
        if(!n)
        {
            printf("0\n");
           continue;
        }
        printf("%d",b[0]);
        for(i = 1;i<n;i++)
        printf(" %d",b[i]);
        printf("\n");
    }

    return 0;
}