Harbin Tech 2015 summer Training BNU16488 Easy Task (simple question)

A-Easy TaskTime limit:$ MS Memory Limit:65536KB 64bit IO Format:%lld &am P %llu SubmitStatusPracticeZOJ 2969

Description

Calculating the derivation of a polynomial is an easy task. Given a function f (x), we use the (f (x)) ' to denote its derivation. We use the 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's easy to prove the derivation a polynomial is also a polynomial.

Here comes the problem, given a polynomial f (x) with non-negative coefficients, can-write a program to calculate the D Erivation of it?

Input

Standard input would contain multiple test cases. The first line of the input was a single integert (1 <= T <=) which was the number of the test case S. And it'll be followed byT consecutive test cases.

There is exactly 2 lines in each test case. The first line of all test case was a single line containing an integerN (0 <= n <= 100). The second line contains N + 1 non-negative integers,CN, CN-1, ..., C1, C0, ( 0 <= Ci <=), which is 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
(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 integersCm,Cm-1,... C0.
(3) There is a single space between the integers and no spaces after the last integer.

Sample Input

301023 2 1310 0 1 2

Sample Output

06 230 0 1 Field second kill problem. Not explained.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include < string.h> #include <cctype> #include <string> #include <cmath> #include <vector> #include <stack> #include <queue> #include <map> #include <set>using namespace Std;int main () {    int CNT[10000];    int t;cin>>t;    while (t--)    {        int n;cin>>n;        for (int i=0;i<=n;i++)        {            scanf ("%d", &cnt[i]);        }        if (n==0)        {            cout<<0<<endl;continue;        }        int x=n;        cout<<cnt[0]*x;x--;        for (int i=1;i<n;i++)        {            cout<< "" <<cnt[i]*x;            x--;        }        cout<<endl;    }    return 0;}

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Harbin Tech 2015 summer Training BNU16488 Easy Task (simple question)