1009. Product of Polynomials (25)-PAT甲级真题

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

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• 输入两个多项式
  • k 非0项
  • 指数和系数
• 输出他们的乘积，指数降序，小数保留1位
  • k 非0项
  • 指数和系数

分析

用double数组存储输入，ans数组存储结果，并将ans数组输出

• 输入一个多项式，用arr数组接收
• 输入第二个多项式，输入的时计算该项与第一个多项式各项的乘积ans[j + a] += arr[j] * b;
• 统计非零项的个数cnt
• 输出结果

#include 

using namespace std;

int main() {
    int n1, n2, a, cnt = 0;
    cin >> n1;
    double b, arr[1001] = {0.0}, ans[2001] = {0.0};
    for (int i = 0; i < n1; i++) {
        cin >> a >> b;
        arr[a] = b;
    }
    cin >> n2;
    for (int i = 0; i < n2; i++) {
        cin >> a >> b;
        for (int j = 0; j < 1001; j++)
            ans[j + a] += arr[j] * b; //x^(j+a)的系数=(x^j的系数)*(x^a的系数)
    }
    //统计非零项
    for (int i = 2000; i >= 0; i--) {
        if (ans[i] != 0)
            cnt++;
    }
    cout << cnt;
    for (int i = 2000; i >= 0; i--)
        if (ans[i] != 0.0)
            printf(" %d %.1f", i, ans[i]);
    return 0;
}