PAT甲级——A1009 Product of Polynomials

概述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 informati

This time,you are supposed to find @H_403_11@A×B where @H_403_11@A and @H_403_11@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:

@H_403_11@K @H_403_11@N?1?? @H_403_11@a?N?1???? @H_403_11@N?2?? @H_403_11@a?N?2???? ... @H_403_11@N?K?? @H_403_11@a?N?K????

where @H_403_11@K is the number of nonzero terms in the polynomial, @H_403_11@N?i?? and @H_403_11@a?N?i???? (,) are the exponents and coefficIEnts,respectively. It is given that 1, 0.

Output Specification:

For each test case you should output the product of @H_403_11@A and @H_403_11@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:

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         else35             printf(" %d %.1f",ptr->first,ptr->second);36     }37     cout << endl;38 39     return 0;40 }

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