1009. Product of Polynomials (25)

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

//定番

#include<stdio.h>#include<iostream>#include<string>using namespace std;#define SIZE 2001int main(){int k1, k2, k;double p1[SIZE], p2[SIZE], p[SIZE];for(int i = 0; i < SIZE; i++){p1[i] = p2[i] = p[i] =0;}cin>>k1;for(int i = 0; i < k1; i++){cin>>k;cin>>p1[k];}//for(int i = 0; i < k1; i++)//{//cout<<p1[i]<<" ";//}cin>>k2;for(int i = 0; i < k2; i++){cin>>k;cin>>p2[k];}for(int i = 0; i < SIZE; i++){if(p1[i] != 0){for(int j = 0; j < SIZE; j++){if(p2[j] != 0){double a = p1[i]*p2[j];int b = i+j;p[b] += a;}}}}int cnt = 0; for(int i = 0; i < SIZE; i++){if(p[i] != 0)cnt++;}cout<<cnt;for(int i = SIZE-1; i >= 0; i--){if(p[i] != 0){printf(" %d %.1lf", i, p[i]);}}return 0;}