PAT-A-1009. Product of Polynomials (25)

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 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

#include<iostream>#include<cstdio>using namespace std;int main(){double p2[2002] = { 0 };double p[1001] = { 0 };double m;int n;int k;int count = 0;cin >> k;for (int i = 0; i < k; i++){cin >> n >> m;p[n] = m;}cin >> k;for (int i = 0; i < k; i++){cin >> n >> m;for (int j=0; j < 1001; j++){if (p[j] != 0){p2[j + n] += p[j] * m;}}}for (int j=0; j < 2002; j++){if (p2[j] != 0){count++;}}cout << count;for (int i = 2001; i >= 0; i--){if (p2[i] != 0){cout << " " << i << " ";printf("%.1lf", p2[i]);}}system("pause");return 0;}

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