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 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 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

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int main(){
 5     int K, n, count = 0;
 6     double a, poly1[1001] = {0}, re[2002] = {0};
 7     scanf("%d", &K);
 8     for(int i = 0; i < K; i++){
 9         scanf("%d%lf", &n, &a);
10         poly1[n] = a;
11     }
12     scanf("%d", &K);
13     for(int i = 0; i < K; i++){
14         scanf("%d%lf", &n, &a);
15         for(int j = 0; j < 1001; j++)
16             re[n + j] = re[n + j] + poly1[j] * a;
17     }
18     for(int i = 2001; i >= 0; i--){
19         if(re[i] != 0)
20             count++;
21     }
22     printf("%d", count);
23     for(int i = 2001; i >= 0; i--){
24         if(re[i] != 0)
25             printf(" %d %.1lf", i, re[i]);
26     }
27     cin >> K;
28     return 0;
29 }
View Code

 

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。

posted @ 2018-01-17 14:09  ZHUQW  阅读(120)  评论(0)  编辑  收藏  举报