A1009. Product of Polynomials

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 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

 #include<cstdio>
 #include<iostream>
 using namespace std;
 int main(){
     int K, n, count = ;
     double a, poly1[] = {}, re[] = {};
     scanf("%d", &K);
     for(int i = ; i < K; i++){
         scanf("%d%lf", &n, &a);
         poly1[n] = a;
     }
     scanf("%d", &K);
     for(int i = ; i < K; i++){
         scanf("%d%lf", &n, &a);
         for(int j = ; j < ; j++)
             re[n + j] = re[n + j] + poly1[j] * a;
     }
     for(int i = ; i >= ; i--){
         if(re[i] != )
             count++;
     }
     printf("%d", count);
     for(int i = ; i >= ; i--){
         if(re[i] != )
             printf(" %d %.1lf", i, re[i]);
     }
     cin >> K;
     return ;
 }

总结：

1、指数为1000的多项式乘法，结果最高为2000次幂。

2、为减少时间复杂度，可以将第一个多项式存储，第二个不存。第二个多项式边读入边直接遍历poly1并做乘法。还可以将两个多项式的系数与指数分别开数组存下来以减小复杂度。