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 N1 a**N1 N2 a**N2 … N**K aNK
where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N**K<⋯<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 结尾无空行
Sample Output:
3 3 3.6 2 6.0 1 1.6 结尾无空行
#include <iostream> using namespace std; int main() { int N1, N2; cin >> N1; double arr[1005] = {0.0}, ans[2005] = {0.0}; int z; double x; for (int i = 0; i < N1; i++) { cin >> z >> x; arr[z] = x; } cin >> N2; for (int i = 0; i < N2; i++) { cin >> z >> x; for (int j = 0; j < 1005; j++) { if (arr[j]) { ans[z + j] += arr[j] * x; } } } int count = 0; for (int i = 2004; i >= 0; i--) { if (ans[i]) { count++; } } cout << count; for (int i = 2004; i >= 0; i--) { if (ans[i]) { printf(" %d %.1f", i, ans[i]); } } system("pause"); return 0; }
多项式相乘，与1002的思路一致。
