PTA 1002 A+B for Polynomials

问题描述：

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​​​​ (,) are the exponents and coefficients, respectively. It is given that 1，0.

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

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这题测试数据魔鬼！魔鬼！魔鬼！
同志们，作为多项式看待的时候，常数0不应该是算有1项且N1和aN1均为0吗？！我考虑了这种情况，结果有一个结果迟迟不对，哪料到把这个删了就对了，天理难容！
代码是同学发给我改的，所以比较丑。同学代码写的比较臃肿，缩进也挺不舒服的，凑合着看吧。我改的时候都没搞成tab缩进，现在自然也懒得弄了，反正AC了。

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代码：

 #include<iostream>
 #include<iomanip>
 #include<cmath>
 using namespace std;
 int atotal[];
 int ktotal;
 float btotal[];

 void mix(int na,int a[],int nb,int b[],float c[],float d[])
 {
  int ia=,ib=;
  while(ia<na&&ib<nb)
  {
   if (a[ia]>b[ib])
   {
    atotal[ktotal]=a[ia];
    btotal[ktotal]=c[ia];
    ia++;
   }
   else if (a[ia]<b[ib])
   {
    atotal[ktotal]=b[ib];
    btotal[ktotal]=d[ib];
    ib++;
   }
   else
   {
    atotal[ktotal]=a[ia];
    btotal[ktotal]=c[ia]+d[ib];
    ia++;ib++;
   }
   if (!(btotal[ktotal]>-0.05&&btotal[ktotal]<0.05)) ktotal++;
  }
  while(ia<na)
  {
   atotal[ktotal]=a[ia];
   btotal[ktotal]=c[ia];
   if (!(btotal[ktotal]>-0.05&&btotal[ktotal]<0.05)) ktotal++;
   ia++;
  }
  while(ib<nb)
  {
   atotal[ktotal]=b[ib];
   btotal[ktotal]=d[ib];
   if (!(btotal[ktotal]>-0.05&&btotal[ktotal]<0.05)) ktotal++;
   ib++;
  }
 }

 int main()
 {
  int k1,k2;
  int a1[],a2[];
  float b1[],b2[];
  cin>>k1;
  for (int i=;i<k1;i++){
   cin>>a1[i];
   cin>>b1[i];
  }
  cin>>k2;
  for (int i=;i<k2;i++)
  {
   cin>>a2[i];
   cin>>b2[i];
  }
  mix(k1,a1,k2,a2,b1,b2);
  cout<<ktotal;
  cout.precision ();
  cout.setf(ios::fixed | ios::showpoint );
  for (int i=;i<ktotal;i++){
      cout << " " << atotal[i] << " " << round(*btotal[i])/10.0;
  }
  return ;
 }