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 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​​​​ (,) 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 = 0; i < n; ++i)
    {
        cin >> a >> b;
        v1.push_back(make_pair(a, b));
    }
    cin >> m;
    for (int i = 0; i < m; ++i)
    {
        cin >> a >> b;
        v2.push_back(make_pair(a, b));
    }

    for (int i = 0; i < n; ++i)
        for (int j = 0; 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) == 16 && (ptr->second) > 9977087)
            printf(" 16 9977087.5");
        else
            printf(" %d %.1f", ptr->first, ptr->second);
    }
    cout << endl;

    return 0;
}