1009 Product of Polynomials (25分)
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 N 1 a N 1 N 2 a N 2 ... N K a N K where K is the number of nonzero terms in the polynomial, N i and a N i (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N K <⋯<N 2 <N 1 ≤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
时间限制: 400 ms
内存限制: 64 MB
代码长度限制: 16 KB
思路
这道题是一个多项式的乘法，考察map的使用和如何定义按照指定规则排序的map，
利用map进行遍历即可，也可以使用数组的方法，注意系数为0的情况要被删除！
**这题坑在测试点1,有系数为0的项。就是多项式的每一项乘另外多项式的每一项然后相加,然后合并同类项之后为0 ！！！
1 #include <iostream>
2 #include <stdio.h>
3 #include <map>
4 #include <algorithm>
5 #include <iterator>
6
7
8 using namespace std;
9
10 double a[30];
11 double b[30];
12 map<int, double, greater<int> > mp;
13
14 int main()
15 {
16 int ka, kb;
17 cin >> ka;
18 ka = 2*ka;
19 for(int i = 1; i <= ka; i++)
20 {
21 cin >> a[i];
22 }
23
24 cin >> kb;
25 kb = 2*kb;
26 for(int i = 1; i <= kb; i++)
27 {
28 cin >> b[i];
29 }
30
31
32 for(int i = 1; i <= ka; i+=2)
33 {
34 for(int j = 1; j <= kb; j+=2)
35 {
36 mp[a[i]+b[j]] += a[i+1]*b[j+1];
37 if(mp[a[i]+b[j]] == 0) // 合并后系数有可能为0，此时要删去
38 mp.erase(a[i]+b[j]);
39 }
40 }
41
42 int Size = mp.size();
43 printf("%d", Size);
44 for(auto iter = mp.begin(); iter != mp.end(); ++iter)
45 {
46 printf(" %d %.1f", iter->first, iter->second);
47 }
48
49
50 return 0;
51 }
