甲级1002. A+B for Polynomials

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 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




分析：将系数相同的项，变量相加，就ok了，但是注意一个地方，系数可以为0，但是变量不可以为0（哭了，不知道有这种规定），所以输出时把变量为0的剔除掉

代码如下：

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

struct term {
    int times;
    double num;
};

bool cmp(term a, term b) { return a.times > b.times; }

unordered_map<int, int> mapp;
vector<term> poly;

int main()
{
    int a, b, exp;
    double coe;
    scanf("%d", &a);
    for (int i = 0; i < a; i++)
    {
        scanf("%d %lf", &exp, &coe);
        auto it = mapp.find(exp);
        if (it != mapp.end())
        {
            //在map中找到
            poly[it->second].num += coe;
        }
        else
        {
            //没找到
            int count = poly.size();
            mapp[exp] = count;        //map中存入该键值对
            term temp = { exp,coe };
            poly.push_back(temp);
        }
    }
    scanf("%d", &b);
    for (int i = 0; i < b; i++)
    {
        scanf("%d %lf", &exp, &coe);
        auto it = mapp.find(exp);
        if (it != mapp.end())
        {
            //在map中找到
            poly[it->second].num += coe;
        }
        else
        {
            //没找到
            int count = poly.size();
            mapp[exp] = count;        //map中存入该键值对
            term temp = { exp,coe };
            poly.push_back(temp);
        }
    }
    sort(poly.begin(), poly.end(), cmp);
    int count=0;
    for (auto it = poly.begin(); it != poly.end(); it++)
    {
        if (it->num != 0)    count++;
    }
    cout << count;
    for (auto it = poly.begin(); it != poly.end(); it++)
    {
        if(it->num!=0)
            printf(" %d %.1lf", it->times, it->num);
    }
    return 0;
}

这里用了一个vector和一个 map，因为一开始看到这个想到的就是用数组，而寻找多项式的各个项对应的位置又想到了用字典。

看有人用的是map而不是unoder_map，然后利用map的自动键值从小到大排序来实现系数的有序。我使用数组来记录多项式，而他的做法用字典的键值对来记录多项式。

这里贴上他的做法，我觉得非常ok。

#include<cstdio>
#include<map>
using namespace::std;
map<int,double> s;
int main()
{
    int i1,t1;
    double t2;
    for(int z=0;z<2;z++)
    {
        scanf("%d",&i1);
        for(int i=0;i<i1;i++)
        {
            scanf("%d%lf",&t1,&t2);
            if(s.count(t1)==0)
                s[t1]=t2;
            else
                s[t1]+=t2;
        }
    }
    i1=0;
    for(map<int,double>::const_iterator m_it=s.begin();m_it!=s.end();m_it++)
    {
        if(m_it->second!=0.0&&m_it->second!=-0.0)
            i1++;
    }
    printf("%d",i1);
    for(map<int,double>::reverse_iterator m_it=s.rbegin();m_it!=s.rend();m_it++)
    {
        if(m_it->second!=0.0&&m_it->second!=-0.0)
            printf(" %d %.1lf",m_it->first,m_it->second);
    }
    printf("
");
    return 0;
}

--------------------- 
作者：漂流瓶jz 
来源：CSDN 
原文：https://blog.csdn.net/qq278672818/article/details/54563738 
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