PAT甲级练习题A1002. A+B for Polynomials

题目描述

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 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

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

题目解析

第一种我使用两个链表储存两个多项式，编程过程中出现两个问题，一个是输出的格式不对；另外一个是在循环中continue使用的不对，导致错误的跳过了循环体中后面的过程，这个找了一个晚上才发现，一直有两个过不了。
第二种是使用了数组的下标代表次数，因为次数小于1000，所以可以直接用简单的方法。而且只要一个数组就OK了。所以做题目的时候要注意条件，选择合适的简单的方法解决问题。

代码

第一种

//错误的continueif ( N1==N2 )    {      ++i, ++j;      if (an1 + an2 < 0.1)      {        continue;      }      element ele(N1, an1 + an2);      out.push_back(ele);    }

改正后

#include<iostream>#include<vector>#include<cmath>using namespace std;class element{public:  int expo;  double coeff;  element(int ex = 0, double co = 0) :expo(ex), coeff(co) {};};int main(){  int K, N;  double an;  vector<vector<element> > poly(2);  vector<element> out;  for (int i = 0; i < 2; ++i)  {    cin >> K;    for (int j = 0; j < K; ++j)    {      cin >> N >> an;      element ele(N, an);      poly[i].push_back(ele);    }  }  for (int i = 0, j = 0; i < poly[0].size() && j < poly[1].size();)  {    int N1 = poly[0][i].expo, N2 = poly[1][j].expo;    double an1 = poly[0][i].coeff, an2 = poly[1][j].coeff;    //原来的错误在这里    if ( N1==N2 )    {      ++i, ++j;      if (an1 + an2 != 0) //这里我觉得可以考虑下浮点数的边界问题，不过通过了      {        element ele(N1, an1 + an2);        out.push_back(ele);      }    }    else if (N1 > N2)    {      element ele(N1, an1);      out.push_back(ele);      ++i;    }    else    {      element ele(N2, an2);      out.push_back(ele);      ++j;    }    if (i == poly[0].size())    {      for (int k = j; j < poly[1].size(); ++j)      {        out.push_back(poly[1][j]);      }    }    else if (j == poly[1].size())    {      for (int k = i; i < poly[0].size(); ++i)      {        out.push_back(poly[0][i]);      }    }  }  cout << out.size();  for (int i = 0; i < out.size(); ++i)  {    printf(" %d %.1f",out[i].expo,out[i].coeff);  }  cout << endl;  system("pause");  return 0;}

第二种

#include<iostream>#include<vector>using namespace std;int maxexp = 1001;vector<double> poly(maxexp, 0);int main(){  int K, N;  double an;  int i = 2;  while (i--)  {    cin >> K;    while (K--)    {      cin >> N >> an;      poly[N] += an;    }  }  int cnt = 0;  for (int i = maxexp - 1; i >= 0; --i)  {    if (poly[i] != 0)    {      ++cnt;    }  }  cout << cnt;  if (cnt != 0)  {    for (int i = maxexp - 1; i >= 0; --i)    {      if (poly[i] != 0)      {        printf(" %d %.1f", i, poly[i]);      }    }  }  cout << endl;  system("pause");  return 0;}