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​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

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

 

多项式求和

输入为两行     一个多项式的项数k 然后给出k组  e次幂 和 a系数

输出一行 同样　　 多项式项数k   k组  e次幂 和 a系数

本身不难,但是要注意 当同幂次系数相加 和为0 时,需要删除这一项

这里用的是map<int,double>int 是幂,double是值 , 直接删除  key-value 键值对,  用map.erase(iterator);参数为指向key的迭代器

测试点 3 4 5 6 当最终结果没有数据的时候, 只输出项数 0 即可, 不需要输出多项式 

 

附一个来自其他同学的测试用例https://blog.csdn.net/han_hhh/article/details/106338514

input
0
0
output
0
 
input
1 1000 1.5
0
output
1 1000 1.5
 
input
1 1000 1.5
1 1000 -1.5
output
0
 
input
2 1000 1.5 500 1.5
1 1000 -1.5
output
1 500 1.5

 

 AC后 去除了重复多余代码的版本

#include <iostream>
#include <vector>
#include <map>
#include <iomanip>
using namespace std;
void getPolynomial(map<int,double> &a){
    int k,ni;
    double ai;
    cin >> k;
    map<int,double>::iterator it;
    for(int i=0;i<k;i++){
        cin >> ni >> ai;
        a[ni]+=ai;
        if(a[ni]==0)
        {
            it=a.find(ni);
            a.erase(it);
        }
    }
}
int main(){
    map<int,double>::iterator it;
    map<int,double> polynomial;
    vector<int> ni;
    vector<double> ai;
    getPolynomial(polynomial);
    getPolynomial(polynomial);
    for(it=polynomial.begin();it!=polynomial.end();it++){
        ni.push_back(it->first);
        ai.push_back(it->second);
    }
    cout << ai.size();
    cout.setf(ios::fixed);
    for(int i=ni.size()-1;i>=0;i--){
        cout << " "<< ni[i]<<" "<< setprecision(1) <<ai[i];
    }
    return 0;
}

 

 

 第一次提交

//多项式求和
#include <iostream>
#include <vector>
#include <map>
#include <iomanip>
using namespace std;
void getPolynomial(map<int,double> &poly){
    int k,ni;
    double ai;
    cin >> k;
    for(int i=0;i<k;i++){
        cin >> ni >> ai;
        poly[ni]=ai;
    }
}
void aplusb(map<int,double> &a,map<int,double>&b){
    map<int,double>::iterator it;
    map<int,double>::iterator itb;
    for(it=a.begin();it!=a.end();it++){
        itb=b.find(it->first);
        b[it->first]+=it->second;
        if(b[it->first]==0)
        {
            b.erase(itb);
        }
    }
}
int main(){
    map<int,double>::iterator it;
    map<int,double> polynomial1;
    map<int,double> polynomial2;
    map<int,double> result;
    vector<int> ni;
    vector<double> ai;
    getPolynomial(polynomial1);
    getPolynomial(polynomial2);
    aplusb(polynomial1, result);
    aplusb(polynomial2, result);
    for(it=result.begin();it!=result.end();it++){
        ni.push_back(it->first);
        ai.push_back(it->second);
    }
    cout << ai.size();
    cout.setf(ios::fixed);
    for(int i=ni.size()-1;i>=0;i--){
        cout << " "<< ni[i]<<" "<< setprecision(1) <<ai[i];
    }
    return 0;
}