PAT甲级1009多项式相乘

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 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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

题目大意

给出两个多项式A和B，求A*B的结果

注意

错因：c开了1010。应该开2000+
方法：用数组下标作为幂次，用数组元素存储系数

//
//  PAT 甲 1009多项式乘积.cpp
//  Fushi
//
//  Created by H C on 2021/2/5.
//  题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805509540921344

// 错因：c开了1010。应该开2000+
// 方法：用数组下标作为幂次，用数组元素存储系数

#include <stdio.h>
using namespace std;

double a[1010]={0};
double b[1010]={0};
double c[2010]={0};

int main()
{
    int K1,K2;
    scanf("%d",&K1);
    while (K1--) {
        int N;
        double A;
        scanf("%d %lf",&N,&A);
        a[N] = A;
    }
    scanf("%d",&K2);
    while (K2--) {
        int N;
        double B;
        scanf("%d %lf",&N,&B);
        b[N] = B;
    }
    int count=0;
    for (int i=0; i<1010; i++)
    {
        if (a[i]!=0) {
            for(int j=0; j<1010; j++)
            {
                if(b[j]!=0) {c[i+j] += a[i]*b[j];}
            }
        }
    }
    for (int i=2010; i>=0; i--) {
        if(c[i]!=0) count++;
    }
    printf("%d",count);
    for (int i=2010; i>=0; i--) {
        if (c[i]!=0) {
            printf(" %d %.1lf",i,c[i]);
        }
    }
}