1009. Product of Polynomials

题目是pat甲级

思路：用a和b两个数组分别保存两个多项式，下标表示指数，值表示系数。用c表示乘积。 计算公式：c[i+j]+=a[i]*b[j]

目的：多项式存储  多项式乘积 数组

题目描述：

This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input

2 1 2.4 0 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

参考代码：

#include <iostream>#include <iomanip>using namespace std;#include <string.h>int main(){  double a[1001],b[1001],c[2001];  int k;  while(cin>>k)  {    memset(a,0,sizeof(double)*1001);    memset(b,0,sizeof(double)*1001);    memset(c,0,sizeof(double)*2001);    int i,j;    for(i=0;i<k;i++)    {      double bi;      int e;      cin>>e>>bi;      a[e]=bi;    }    cin>>k;    for(i=0;i<k;i++)    {      double bi;      int e;      cin>>e>>bi;      b[e]=bi;    }    for(i=0;i<1001;i++)      for(j=0;j<1001;j++)        c[i+j]+=a[i]*b[j];    int count=0;    for(i=0;i<2001;i++)      if(c[i]!=0)        count++;    cout<<count;    for(i=2000;i>=0;i--)      if(c[i]!=0)        cout<<" "<<i<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<c[i];    cout<<endl;  }  return 0;}