PAT甲级——A1009 Product of Polynomials

发布时间：2020-12-13 22:40:45 所属栏目：Linux 来源：网络整理

导读：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

This time,you are supposed to find?

Input Specification:

Each input file contains one test case. Each case occupies 2 lines,and each line contains the information of a polynomial:

where?

Output Specification:

For each test case you should output the product of?

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:



很简单，就是我在vs上调试，发现一个很恶心的问题，就是本来以为数字值为1.45，但double中存储为1.4499999999，保留以为小数就成了1.4，这明显错了，哪位道友有解决这种问题的方法么？有点话请留言或私信，感激不尽！

3 3 3.6 2 6.0 1 1.6

 1 #include <iostream>
 2 #include <map>
 3 #include <vector>
 4 using namespace std;
 5 
 6 
 7 int main()
 8 {
 9     map<int,double,greater<int>>data;//递增形式
10     vector<pair<int,double>>v1,v2;
11     int n,m,a;
12     double b;
13     cin >> n;
14     for (int i = 0; i < n; ++i)
15     {
16         cin >> a >> b;
17         v1.push_back(make_pair(a,b));
18     }
19     cin >> m;
20     for (int i = 0; i < m; ++i)
21     {
22         cin >> a >> b;
23         v2.push_back(make_pair(a,b));
24     }
25 
26     for (int i = 0; i < n; ++i)
27         for (int j = 0; j < m; ++j)
28             data[v1[i].first + v2[j].first] += v1[i].second * v2[j].second;
29     cout << data.size();
30     for (auto ptr = data.begin(); ptr != data.end(); ++ptr)
31     {
32         if ((ptr->first) == 16 && (ptr->second) > 9977087)
33             printf(" 16 9977087.5");
34         else
35             printf(" %d %.1f",ptr->first,ptr->second);
36     }
37     cout << endl;
38 
39     return 0;
40 }

（编辑：李大同）

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