1013: Polynomial Multiplication

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本文介绍了一种解决多项式乘法问题的算法实现方法，输入为两个多项式的系数，输出为这两个多项式相乘的结果。文章通过示例展示了如何进行输入输出，并提供了一个C语言的实现代码。

1013: Polynomial Multiplication

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	4043	692	Standard

A polynomial is a sum of terms, where each term is a coefficient multiplied by the variable x raised to some nonnegative integer power. The largest such power is referred to as the degree of the polynomial. For example: x^3+x^2+x+1, x^3+x, x^2, and 1 are all polynomials.

Input

A positive integer denoting the number of cases.
(The degree of polynomial 1) followed by the value of each coefficient for polynomial 1.
(The degree of polynomial 2) followed by the value of each coefficient for polynomial 2.
No polynomial will have degree greater than five. All coefficients will be greater than or equal to zero and less than 65536. Each polynomial will be on a separate line. A blank line will separate each case.

Output

The product of polynomial 1 and polynomial 2. The caret symbol (^) should be used to denote exponentiation. That is, x^2 means x*x. But, instead of x^1, just use x, and instead of x^0, use nothing. Terms where the coefficient is 0 must be omitted.

Example

In this example we multiply the following two sets of polynomials:
x^4+2x^3+x^2+5x and x^3+2x^2+1
x^3+x+1 and 2x^3+x+2

Input

2
4 1 2 1 5 0
3 1 2 0 1

3 1 0 1 1
3 2 0 1 2

Output

x^7 + 4x^6 + 5x^5 + 8x^4 + 12x^3 + x^2 + 5x
2x^6 + 3x^4 + 4x^3 + x^2 + 3x + 2

#include<stdio.h>
void reverse(int *p,int n)
{
int i,j,t;
for(i=0,j=n;i<j;i++,j--)
{
  t=*(p+i);
  *(p+i)=*(p+j);
  *(p+j)=t;
}
}
int main()
{
int n,t,a,b,i,j,s;
scanf("%d",&n);
for(t=0;t<n;t++)
{
  int f[30]={0},g[30]={0},z[30]={0};

  scanf("%d",&a);
  for(i=0;i<=a;i++)
   scanf("%d",f+i);
  reverse(f,a);
  scanf("%d",&b);
  for(j=0;j<=b;j++)
   scanf("%d",g+j);
  reverse(g,b);
  for(i=0;i<=a;i++)
   for(j=0;j<=b;j++)
    z[i+j]+=f[i]*g[j];
  s=a+b;
int al = 0;
while(s >= 0)
{
  if(z[s] != 0)
  {
   if(al == 0)
   {
    al = 1;
   }
   else
    printf(" + ");
   if(z[s]!=1 || s==0)
    printf("%d",z[s]);
   if(s != 0)
   {
    printf("x");
    if(s != 1)
     printf("^%d", s);
   }
  }
  --s;
}
if(al == 0) printf("0");
printf("/n");
}
return 0;
}