zoj 2969 Easy Task

ZOJ Problem Set - 2969 Easy Task

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Time Limit: 1 Second      Memory Limit: 32768 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

3 0 10 2 3 2 1 3 10 0 1 2

 

Sample Output

0 6 2 30 0 1

 

Author: CAO, Peng

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Source: The 5th Zhejiang Provincial Collegiate Programming Contest Submit    Status

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// Wpl // 1786815 2009-03-12 20:30:35 Accepted  2969 C++ 20 184 吴 // ZOJ 2969 Easy Task 简单积分题 08年浙江省大学生程序设计竞赛 #include  < iostream > using   namespace  std; int  C[ 105 ]; int  main() {      int  t;     scanf( " %d " , & t);      while (t -- )     {          int  n,i;         scanf( " %d " , & n);          for (i = 0 ;i <= n;i ++ )             scanf( " %d " , & C[i]);          for (i = 0 ;i <= n;i ++ )             C[i] = C[i] * (n - i);          for (i = 0 ;i < n - 1 ;i ++ )             printf( " %d  " ,C[i]);         printf( " %d\n " ,C[i]);     }      return   0 ; }

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/12/1454655.html

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