PAT 1009 Product of Polynomials

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

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
#include <vector>
#include <strstream>
#include <map>

using namespace std;
struct Node
{
    int x;
    double y;
}a[15],b[15];
double tag[2005];
int n1,n2;
int main()
{
    scanf("%d",&n1);
    for(int i=1;i<=n1;i++)
        scanf("%d%lf",&a[i].x,&a[i].y);
    scanf("%d",&n2);
    for(int i=1;i<=n2;i++)
        scanf("%d%lf",&b[i].x,&b[i].y);
    memset(tag,0,sizeof(tag));
    for(int i=1;i<=n1;i++)
    {
        for(int j=1;j<=n2;j++)
        {
            tag[a[i].x+b[j].x]+=a[i].y*b[j].y;
        }
    }
    int num=0;
    for(int i=2000;i>=0;i--)
    {
        if(tag[i]!=0)
            num++;
    }
    if(num==0)
    {
        printf("%d
",num);
        return 0;
    }
    else
        printf("%d ",num);
    int cnt=0;
    for(int i=2000;i>=0;i--)
    {
        if(tag[i]!=0)
        {
            cnt++;
            if(cnt==num)
                printf("%d %.1f
",i,tag[i]);
           else
                printf("%d %.1f ",i,tag[i]);
        }
    }
    return 0;

}