• ZOJ Problem Set 2969 Easy Task


    Easy Task

    Time Limit: 1 Second      Memory Limit: 32768 KB

    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, CNCN-1, ..., C1C0, ( 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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  • 原文地址:https://www.cnblogs.com/malloc/p/2097303.html
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