ZOJ 2969 Easy Task

E - 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

0
6 2
30 0 1

题意就是求导……

没啥好说的

#include<iostream>  
#include<string.h>  
#include<stdio.h>  
#include<ctype.h>  
#include<algorithm>  
#include<stack>  
#include<queue>  
#include<set>  
#include<math.h>  
#include<vector>  
#include<map>  
#include<deque>  
#include<list>  
using namespace std;  
int main()
{
    int a;
    cin>>a;
    for(int i=1;i<=a;i++)
    {
        int b;
        cin>>b;
        int p=b+1;
        while(p--)
        {
            int c;
            cin>>c;
            if(b==0)
            {
            printf("0\n"); 
            break;
            }
            if(p==0)
            {
            printf("\n");
            break;
            }
            if(p==1)
            printf("%d",c*p);
            else
            printf("%d ",c*p);
        }
        
    }
    return 0;
}