1009 Product of Polynomials (25 分) PAT甲级1009 模拟多项式相乘巧用STL 我的代码+别人的代码

谁践踏了优雅 2024-02-19 20:52 52阅读 0赞

[1009 Product of Polynomials (25 分)][1009 Product of Polynomials _25]

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 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 and coefficients, respectively. It is given that 1≤K≤10, 0≤NK<⋯<N2<N1≤1000.

#### Output Specification: ####

For each test case you should output the product 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 up to 1 decimal place.

#### Sample Input: ####

2 1 2.4 0 3.2
    2 2 1.5 1 0.5

#### Sample Output: ####

3 3 3.6 2 6.0 1 1.6

题意：多项式相乘

我的思路：用stl里的map，声明3个map<int,double>容器，遍历容器，用第一个容器里的每一项去乘第二个容器的每一项，得到的结果存到C容器里。但是要注意：多项式的系数可能为0，所以要在C容器里去掉系数为0的项。（第一个测试点卡这个）

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<map>
    using namespace std;
    int main()
    {
    	int i,j,n,m;
    	int r=0;
    	double index;
    	map<int,double> a,b;
    	map<int,double,greater<int> >c;  //结果容器按键值从大到小排列 
    	cin>>n;
    	while(n--)
    	{
    		scanf("%d %lf",&r,&index);
    		a.insert(make_pair(r,index));
    	}
    	cin>>m;
    	while(m--)
    	{
    		scanf("%d %lf",&r,&index);
    		b.insert(make_pair(r,index));
    	}
    	for(map<int,double>::iterator it=a.begin();it!=a.end();it++)    // 遍历 
    	{
    		for(map<int,double>::iterator id=b.begin();id!=b.end();id++)
    		{
    			int temp=it->first+id->first;
    			double t=it->second*id->second;
    			if(c.count(temp)!=0)
    			{
    				c[temp]+=t;
    				if(c[temp]==0)     // 如果系数为0，去掉。第一个测试点卡这个 
    				c.erase(temp);
    			}
    			
    			else
    			c.insert(make_pair(temp,t));
    		}
    	}
    	
    	if(c.size()!=0)
    	{
    		cout<<c.size();
    		for(map<int,double>::iterator ir=c.begin();ir!=c.end();ir++)
    		{
    			if(ir->second!=0)
    			printf(" %d %.1lf",ir->first,ir->second);
    		}
    		cout<<endl;
    	}
    	return 0;
    }

## 别人的代码： ##

用浮点型数组存多项式，数组下标是指数，对应的值的系数。

暴力有时也是一种方法

#include<vector>
    #include <sstream>
    #include<cmath>
    #include<iomanip>
    #include<iostream>
    #include <ctype.h>
    #include <stdlib.h>
    #include <algorithm>
     
    using namespace std;
     
    int main()
    {	
    	int n, m;
    	float numa[1001] = {0.0};//存放A中的各项信息，下标是指数，数组的值是对应的系数
    	float numb[1001] = {0.0};
    	float product[2005] = {0.0};//存入乘积的结果
     
    	cin >> n;
    	for (int i = 0; i < n; i++)
    	{
    		int temp;
    		cin >> temp;
    		cin >> numa[temp];
    	}
     
    	cin >> m;
    	for (int i = 0; i < m; i++)
    	{
    		int temp;
    		cin >> temp;
    		cin >> numb[temp];
    	}
    	
    	for (int i = 0; i <1001; i++)//i，j的取值尤其注意，一开始没考虑周全，导致两个测试用例出错
    	{		 
    		for (int j = 0; j <1001; j++)
    		{
    			float coeff = numa[i] * numb[j];
    			int exp = i + j;
    			product[exp] = product[exp]+ coeff;//注意这里的逻辑不要写错了
    		}
    	}
     
    	int count = 0;
    	float t = 0.05;
    	for (int i = 0; i < 2005; i++)//特别注意取值范围
    	{
    		if (abs(product[i])>=t)//这里要特别注意，四舍五入后为0的项，也不能显示，并且要加绝对值才能通过最后一个测试用例，但是我其实不是很明白，他们都是大于0的数，为何要加绝对值，但是不加真的就是通不过第一个测试用例，减5分
    			count++;
    	}
    	cout << count;
    	
    	for (int i = 2004; i >=0; i--)//特别注意取值范围
    	{
    		if (abs(product[i])>= t)
    		{
    			cout << " " << i << " " << setiosflags(ios::fixed) << setprecision(1) << product[i];
    		}			
    	}
     
    	return 0;
    }

[1009 Product of Polynomials _25]: https://pintia.cn/problem-sets/994805342720868352/problems/994805509540921344