A1009. Product of Polynomials (25)

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

// hahaha.cpp : 定义控制台应用程序的入口点。
//

#include <stdafx.h>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <cstdio>
#include <set>
#include <algorithm>
#include <string.h>
using namespace std;

const int maxn=2010;
double a[maxn]={0};
double b[maxn]={0};
double c[maxn]={0};

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        {
        int p;
        double q;
        scanf("%d %lf",&p,&q);
        a[p]=q;
        }
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        {
        int p;
        double q;
        scanf("%d %lf",&p,&q);
        b[p]=q;
        }
    int count=0;
    for(int i=0;i<maxn;i++)
        {
        for(int j=0;j<=i;j++)
            {
            c[i]=c[i]+a[j]*b[i-j];
            }
        if(c[i]!=0)count++;
        }
    printf("%d",count);
    for(int i=maxn-1;i>=0;i--)
          {
             if(c[i]!=0)
              {
              printf(" %d %.1f",i,c[i]);
              }
          }

    return 0;
}