PAT(甲) 1002. A+B for Polynomials (25)

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本文介绍了一种使用C++实现多项式相加的方法。利用标准模板库中的map结构存储多项式的指数和系数，通过输入两个多项式，程序能够计算并输出它们的和。此算法适用于指数不超过1000且项数不多于10的情况。

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1002. A+B for Polynomials (25)

题目地址：1002. A+B for Polynomials (25)

题目描述：

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

输入格式：
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 aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000.
输出格式：
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.

解题方法：
本题可以用map来做，其中指数作为键，系数作为值，这样加减都很容易。

程序：

#include <stdio.h>
#include <map>
#include <algorithm>
using namespace std;

int main()
{
    int K1, K2, key;
    double value;
    map <int, double> polynomial;
    scanf("%d", &K1);
    for (int i = 0; i < K1; i++)
    {
        scanf("%d %lf", &key, &value);
        polynomial[key] += value;
    }
    scanf("%d", &K2);
    for (int i = 0; i < K2; i++)
    {
        scanf("%d %lf", &key, &value);
        polynomial[key] += value;
        if (polynomial[key] == 0)   // 说明变成0项了
            polynomial.erase(key);
    }
    map <int, double>::reverse_iterator it;
    printf("%d", polynomial.size());
    for (it = polynomial.rbegin(); it != polynomial.rend(); it++)
        printf(" %d %.1lf", it->first, it->second); // 降序打印

}