PAT 甲级练习 1002

1002 A+B for Polynomials

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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

For each test case you should output the sum 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 to 1 decimal place.
Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

#include <iostream>#include <map>#include <iomanip>using namespace std;int main() {    bool insertion; // find the insertion point    int K1, K2, expo;    double coef;    map<int, double> result;    cin >> K1;    for(int i=0; i<K1; ++i){        cin >> expo >> coef;        result[expo] = coef;    }    cin >> K2;    for(int i=0; i<K2; ++i){        cin >> expo >> coef;        if(result.find(expo) == result.end())   // not found            result[expo] = coef;        else                                    // found            if(result[expo] + expo == 0)                result.erase(result.find(expo));            else                result[expo] += expo;    }    if(result.empty())  cout << 0 << endl;  // no element    else{   // element exist        cout << result.size() << ' ';        int i = 0, lmt = result.size();        for(auto it=result.rbegin(); it!=result.rend(); ++it){            cout << (*it).first << ' ' << setprecision(1) << fixed << (*it).second;            cout << ((++i == lmt) ? '\n':' ');        }    }    return 0;}

Pay attention that when using map.erase(arg), the arg must be an iterator but not a key.