PAT甲级真题 1009 Product of Polynomials (25分) C++实现（使用map，注意测试点0的坑）

题目

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 (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=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

思路

用数组分别存储第一个多项式的指数和系数，读入第二个多项式时，每读一组指数、系数，就与第一个多项式各项做运算。

结果存储在map里，以指数为key，以系数为value（value默认初始值为0，直接累加即可）。若出现系数为0的项，则将其从map中删除。

最后，map.size()即是非0系数项个数，倒序遍历map中的key和value即为按指数降序排列的结果（map自动按key升序排列）。

注意：

1. 结果要保留1位小数；
2. 可能出现系数为0的情况（题目对系数的输入没有限制），需将其排除在外，否则测试点0无法通过。

代码

#include 
#include 
#include 
#include 
using namespace std;

int main(){
    map<int, double> mmap;

    int k1, k2;
    cin >> k1;
    vector<int> exp(k1);
    vector<double> coef(k1);
    for (int i=0; i<k1; i++){
        cin >> exp[i] >> coef[i];
    }

    cin >> k2;
    for (int i=0; i<k2; i++){
        int tempExp;
        double tempCoef;
        cin >> tempExp >> tempCoef;
        for (int j=0; j<k1; j++){
            int newExp = tempExp + exp[j];
            mmap[newExp] += (tempCoef * coef[j]);
            if (mmap[newExp]==0){
                mmap.erase(newExp);
            }
        }
    }

    cout << mmap.size();
    for (auto it=mmap.rbegin(); it!=mmap.rend(); ++it){
        cout << " " << it->first << " " << setprecision(1) << fixed << it->second;
    }
    cout << endl;
    return 0;
}