ZOJ 2969 模拟 (E)

Easy Task
Time Limit: 2 Seconds      Memory Limit: 65536 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, 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：对方程求导，指数第一项是n，第二项是n-1

题解：直接模拟，这里如果n==1，直接输出0
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define LL long long
#define N 10000
int a
;
#define inf 0x3f3f3f3f
int main(){
#ifdef CDZSC
freopen("i.txt","r",stdin);
#endif
char s
;
int n,t,x;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n+1;i++)
{
scanf("%d",a+i);
}
if(n==0)
{
printf("0");
}
int tt=n;
for(int i=0;i<n;i++)
{
a[i]=a[i]*(tt);
tt--;
}
int fi=1;
for(int i=0;i<n;i++)
{
if(fi)
{
printf("%d",a[i]);
fi=0;
}
else
{
printf(" %d",a[i]);
}
}
printf("\n");
}
return 0;
}