ZOJ2969-Easy Task

Easy Task

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

301023 2 1310 0 1 2

Sample Output

06 230 0 1

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

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，若不止有常数项，常数项就不用输出

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longint main(){    int a[200],n;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0; i<=n; i++)            scanf("%d",&a[i]);        for(int i=0; i<n; i++)        {            if(i!=0) printf(" ");            printf("%d",a[i]*(n-i));        }        if(n==0) printf("0");        printf("\n");    }    return 0;}