PAT1002:A+B for Polynomials

Posted 2020-10-09 0kk470

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

思路

题目要求打印两个多项式相加后的多项式的指数和系数。

用map<指数，系数>这样的关联形式来模拟相加就很简单了，找到对应的指数（键）计算对应的系数（值）就行。
特别注意的情况：指数的系数相加如果为0，那么这个指数和其对应的系数就不用输出了（相消了），而且对应的第一个输出的数字——指数个数也要减1。

代码

#include<iostream>
#include<map>
#include<iomanip>
#include<iterator>
using namespace std;
int main()
{
   int k1,ksum = 0;
   while(cin >> k1)
  {
   map<int,double> sum;  //<指数,系数>
   for(int i = 0;i < k1;i++)
   {
      int n;double a;
      cin >> n >> a;
      sum.insert(pair<int,double>(n,a));
      ksum++;
   }

   int k2;
   cin >> k2;
   for(int i = 0;i < k2;i++)
   {
      int n;double a;
      cin >> n >> a;
      if(sum.count(n) > 0)
      {
        sum[n] += a;
        if(sum[n] == 0)
            ksum--;
      }
      else
      {
         sum.insert(pair<int,double>(n,a));
         ksum++;
      }

   }

   cout << ksum;
   for(map<int,double>::reverse_iterator it = sum.rbegin(); it != sum.rend();it++)
   {
       if(it->second != 0)
       {
         cout << " " << it->first;
         cout <<" "<< fixed << setprecision(1) << it->second;
       }

   }
   cout << endl;
  }
}