A – Easy Task
Time Limit:2000MSMemory Limit:65536KB64bit IO Format:%lld & %llu
ZOJ 2969
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 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,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 integersCm,Cm-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现场秒杀题。不解释。#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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