pat1009

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6
多项式乘法

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<iomanip>
using namespace std;
struct node
{
    int cishu;
    double xishu;
}a[2200];

double a1[2200];
int main()
{

    for(int i=0;i<2200;i++)
    {
        a1[i]=0.0;
    }
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i].cishu>>a[i].xishu;
    }
    int m;
    cin>>m;
    int mx=-1;
    for(int i=0;i<m;i++)
    {
        int q;
        double w;
        cin>>q>>w;
       for(int j=0;j<n;j++)
        {
           int b;
           double c;
           b=q+a[j].cishu;
           c=w*a[j].xishu;
           a1[b]=a1[b]+c;
           if(b>mx)
            mx=b;




        }
    }
        int flag=0;

        for(int i=0;i<=mx;i++)
        {
            if(a1[i]!=0)
            {
                flag++;
            }
        }
        cout<<flag;
         for(int i=mx;i>=0;i--)
         {
             if(a1[i]!=0)
             {
                 cout<<" "<<i;
                 cout<<fixed<<setprecision(1)<<" "<<a1[i];
             }
         }
         printf("
");

    return 0;
}

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