ZOJ2969

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 integerT (1 <= T <= 1000) which is the number of test cases. And it will be followed byT 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 integerN (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 integersC_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

题目就是让你求多项式的一次导数。

注意要对只有常数的多项式做特判断。

#include<bits/stdc++.h>#define clr(x) memset(x, 0, sizeof(x))#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;const int maxn = 100005;int main(){    int T;    cin >> T;    while(T--)    {        int n;        cin >> n;        int a[105];        for(int i = n; i >= 0; i--)            scanf("%d", &a[i]);        if(n == 0)        {            puts("0");            continue;        }        for(int i = n; i >= 1; i--)            printf("%d%c",a[i] * i, i == 1 ? '\n' : ' ');    }}