1009 Product of Polynomials(25 分)(PAT甲級)
阿新 • • 發佈:2019-01-02
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
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
多項式的乘法。
#include <iostream> using namespace std; int main() { int n1, n2, a, cnt = 0; scanf("%d", &n1); double b, arr[1001] = {0.0}, ans[2001] = {0.0}; for(int i = 0; i < n1; i++) { scanf("%d %lf", &a, &b); arr[a] = b; } scanf("%d", &n2); for(int i = 0; i < n2; i++) { scanf("%d %lf", &a, &b); for(int j = 0; j < 1001; j++) ans[j + a] += arr[j] * b; } for(int i = 2000; i >= 0; i--) { if(ans[i] != 0.0) { cnt++; } } printf("%d", cnt); for(int i = 2000; i >= 0; i--) { if(ans[i] != 0.0) { printf(" %d %.1f", i, ans[i]); } } return 0; }