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 Input2 1 2.4 0 3.22 2 1.5 1 0.5Sample Output3 3 3.6 2 6.0 1 1.6
#include <iostream>#include <stdio.h>#include <math.h>using namespace std;double input1[1001];double input2[1001];double result[2003];int main() {int k, exp, i, j;double coef;while(cin>>k) {int count = 0;while(k–) {cin>>exp>>coef;input1[exp] += coef;}cin>>k;while(k–) {cin>>exp>>coef;input2[exp] += coef;}for(i=0; i<=1000; i++) {for(j=0; j<=1000; j++) {result[i+j] += input1[i]*input2[j];}}for(i=0; i<=2000; i++) {if(fabs(result[i]) > 1e-6 ) {count++;}}cout<<count;for(i=2000; i>=0; i–) {if(fabs(result[i]) > 1e-6) {printf(" %d %.1lf", i, result[i]);}}cout<<endl;}return 0;}
，喜欢就该珍惜，珍惜就别放弃。
