浙大pat1009题解

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

32000 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

#include"iostream"
#include "algorithm"
#include <math.h>
#include <iomanip>
#include"string.h"
#include "vector"
using namespace std;
#define max 3000
struct Poly
{
    int index;//指数
    float factor;//系数
};

int main()
{
    vector<double> result;
    int k1,k2;
    cin >> k1;
    vector<Poly> p1(k1);
    for(int i=0;i<k1;i++)
        cin >> p1[i].index >> p1[i].factor;
    cin >> k2;
    vector<Poly> p2(k2);
    for(int i=0;i<k2;i++)
        cin >> p2[i].index >> p2[i].factor;
    result.assign(max+1,0.0);
    int t=0;
    for(int i=0;i<k1;i++)
        for(int j=0;j<k2;j++)
        {
            Poly p;
            p.index = p1[i].index+p2[j].index;
            p.factor = p1[i].factor*p2[j].factor;
            result[p.index] +=p.factor;
        }
    int k=0;

    for(int i=max;i>=0;i--)
    {
        if(fabs(result[i])>1e-6)
        {
            k++;
        }
    }
    cout<<k;
    for(int i=max;i>=0;i--)
    {
        if(fabs(result[i])>1e-6)
        {
            cout<<" "<<i<<" ";
            cout<<setiosflags(ios::fixed);
            cout.precision(1);
            cout<<result[i];
        }
    }
    cout<<endl;
    return 0;
}