【PAT甲级】1002 A+B for Polynomials (25 分)

1002 A+B for Polynomials (25 分)

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 a**N1 N2 a**N2 ... N**K aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N**K<⋯<N2<N1≤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

代码

版本一：

错因：举个例子，当一个多项式中出现10*X³，当读取到这个项式的时候，会把指数3加入到名为con的set容器中,一个多项式中出现了-10 *X³,这时候两个多项式相加，指数为3处的系数是为0的，但在con中并没有将其删除。

所以改动方法即，当v[e]=0时，删除con中对应的那个e

#include<iostream>
#include<vector>
#include<cstdio>
#include<unordered_set>
using namespace std;
int main(){
    vector<double> v(1001,0);//哈希表，下标为代表指数，数组代表对应系数
    unordered_set<int> con;//用来记录有几项非零系数的指数
    for(int i = 0; i < 2; i++){
        int k;
        cin >> k;
        while(k--){
            int e;
            double c;
            cin >> e >> c;
            v[e] += c;
            con.insert(e);
            //第一版错误点，没有考虑到系数相加为0的情况
            if(v[e]==0) con.erase(e);//当相加后系数重新变为0，也要在con中将e删去
        }
    }
    cout << con.size();
    for(int i = 1000; i >= 0; i--){
        if(v[i]!=0){
            printf(" %d %.1f",i,v[i]);//输出格式空格放在前面就不需要那么麻烦

        } 
    }
    return 0;
}