PAT甲级——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 (,) are the exponents and coefficients, respectively. It is given that 1, 0.

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

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！

 #include <iostream>

 #include <map>

 #include <vector>

 using namespace std;

 int main()

 {

     map<int, double, greater<int>>data;//递增形式

     vector<pair<int, double>>v1, v2;

     int n, m, a;

     double b;

     cin >> n;

     for (int i = ; i < n; ++i)

     {

         cin >> a >> b;

         v1.push_back(make_pair(a, b));

     }

     cin >> m;

     for (int i = ; i < m; ++i)

     {

         cin >> a >> b;

         v2.push_back(make_pair(a, b));

     }

     for (int i = ; i < n; ++i)

         for (int j = ; j < m; ++j)

             data[v1[i].first + v2[j].first] += v1[i].second * v2[j].second;

     cout << data.size();

     for (auto ptr = data.begin(); ptr != data.end(); ++ptr)

     {

         if ((ptr->first) ==  && (ptr->second) > )

             printf(" 16 9977087.5");

         else

             printf(" %d %.1f", ptr->first, ptr->second);

     }

     cout << endl;

     return ;

 }

巴特西

PAT甲级——A1009 Product of Polynomials

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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