PAT 1002 A+B for Polynomials（map模拟）

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

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（指数） and coefficients（系数）, 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

题目意思：两个多项式的相加，每一行第一个数k表示非零系数的个数，之后是每个非零系数的指数和系数。

解题思路：可以直接使用数组来模拟，这里我使用map复习一下map的用法。

https://www.cnblogs.com/wkfvawl/p/9387566.html

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<map>
using namespace std;
int main()
{
    int T=;
    int n,k,i;
    double m;
    int cnt=;
    map<int,double>mp;
    map<int,double>::iterator it;//迭代器
    map<int,double>::reverse_iterator rit;//反向迭代器
    while(T--)
    {
        scanf("%d",&k);
        for(i=; i<k; i++)
        {
            scanf("%d%lf",&n,&m);//n为指数,m为系数
            mp[n]+=m;
        }
    }
    for(it=mp.begin();it!=mp.end();it++)//系数为0的
    {
        if(it->second!=)
        {
            cnt++;
            //printf("%lf\n",it->second);
        }
    }
    printf("%d",cnt);
    //map默认从小到大，可以用逆向迭代器逆序输出
    for(rit=mp.rbegin();rit!=mp.rend();rit++)//系数为0的
    {
        if(rit->second!=)
        {
            printf(" %d %.1f",rit->first,rit->second);
        }
    }
    printf("\n");
    return ;
}