PAT(Advanced Level) 1009 Product of Polynomials (25分)

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本题要求实现两个多项式的相乘操作。输入包括两个多项式A和B的系数和指数，输出为A与B相乘的结果，格式与输入相同。题目设定多项式最多包含10项，指数范围在0到1000之间。

PAT(Advanced Level) 1009 Product of Polynomials (25分)

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 $N_1$ $a_{N_1}$ $N_2$ $a_{N_2}$ … $N_K$ $a_{N_K}$ 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≤ $N_K$ <⋯< $N_2$ < $N_1$ ≤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 <bits/stdc++.h>
using namespace std;
int k1, k2;
double A[1005], B[1005], C[1000005];
int main() {
    scanf("%d",&k1);
    for (int i = 0; i < k1; i++){
        int x;double a;
        scanf("%d%lf", &x, &a);
        A[x] = a;
    }scanf("%d", &k2);
    for (int i  = 0; i < k2; i++){
        int x; double a;
        scanf("%d%lf", &x, &a);
        B[x] = a;
    }
    for (int i = 0; i <= 1000; i++)
        for (int j = 0; j <= 1000; j++){
            C[i + j] += A[i] * B[j];
        }
    int cnt = 0;
    for (int i = 0; i <= 1000 * 1000; i++){
       if (C[i] != 0) cnt++;
    }
    printf("%d", cnt);
    for (int i = 1000*1000; i >= 0; i--){
        if (C[i] !=0 ) printf(" %d %.1f", i, C[i]) ;
    }
    return 0;
}