1002. A+B for Polynomials (25)

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本文介绍了一种解决多项式加法问题的算法实现方法，输入包含两个多项式的非零项系数及其指数，输出这两个多项式的和。文章通过C++实现了一个具体的程序案例，详细展示了如何读取输入数据、进行多项式相加并输出结果的过程。

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1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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 <stdio.h>
#include <iostream>
#include <stack>
#include <string.h>
#include <queue>
#include <cmath>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <string>
using namespace std;
typedef long long LL;
bool cmp(pair<int, double> a, pair<int, double> b)
{
    return a.first > b.first;
}
int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt","w",stdout);
    map<int, double> mapp1;
    int n;
    cin >> n;
    for(int i = 0; i < n; i++){
        int esp;
        double coeff;
        cin >> esp >> coeff;
        if(mapp1[esp] != 0){
            mapp1[esp] += coeff;
        }else {
            mapp1[esp] = coeff;
        }
    }
    int m;
    cin >> m;
    for(int i = 0; i < m; i++){
        int esp;
        double coeff;
        cin >> esp >> coeff;
        if(mapp1[esp] != (double)0.0){
            mapp1[esp] += coeff;
        }else {
            mapp1[esp] = coeff;
        }
    }
    vector<pair<int, double>> v1(mapp1.begin(), mapp1.end());
    sort(v1.begin(), v1.end(), cmp);
    
    //有一个坑的地方 当系数互为相反数的时候 是不输出的 
    //所以要重新判断真实的大小 即size
    vector<pair<int, double> > v2;
    for(int i = 0; i < v1.size(); i++){
        if(v1[i].second != (double)0.0){
            pair<int, double> x(v1[i].first, v1[i].second);
            v2.push_back(x);
        }
    }
    cout << v2.size();
    for(int i = 0; i < v2.size(); i++){
        printf(" %d %.1lf", v2[i].first, v2[i].second);
    }

    return 0;
}