PAT A1002 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 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 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

输入输出都是降幂排列
多项式指的是2.4*x^1+3.2*x^0     不是2.4^1+3.2^0   心不在焉的。。。唉。。
系数为0的不输出

第一次麻烦代码：

#include<iostream>
#include<cstdio>
using namespace std;

struct Pol
{
  int n;double a;
}A[11], B[11], C[11];

int main() {
  // freopen("input.txt", "r", stdin);
  int k1, k2;
  cin >> k1;
  for (int i = 0;i<k1;i++) {
    cin >> A[i].n >> A[i].a;
  }
  cin >> k2;
  for (int i = 0;i<k2;i++) {
    cin >> B[i].n >> B[i].a;
  }

  int index = 0, ia = 0, ib = 0;
  while (ia<k1&&ib<k2) {
    if (A[ia].n<B[ib].n) {
      C[index].n = B[ib].n;
      C[index++].a = B[ib++].a;
    }
    else if (A[ia].n == B[ib].n) {
      if(A[ia].a + B[ib].a!=0){
        C[index].n = A[ia].n;
        C[index++].a = A[ia].a + B[ib].a;
      }
      ia++,ib++;
    }
    else {//A[ia].n>B[ib].n
      C[index].n = A[ia].n;
      C[index++].a = A[ia++].a;
    }
  }

  while (ia<k1) {
    C[index].n = A[ia].n;
    C[index++].a = A[ia++].a;//ia++千万别忘了
  }
  while (ib<k2) {
    C[index].n = B[ib].n;
    C[index++].a = B[ib++].a;//ib++千万别忘了   printf大发有时还要快一点
  }

  cout << index;
  for (int i = 0;i<index;i++) {
    if(C[i].a!=0)
    printf(" %d %.1f", C[i].n, C[i].a);
  }
  cout << endl;
  return 0;
}

方法二:
脑子长期不用会锈掉的
本题指数最大1000 完全可以牺牲空间地来做  1000太小了  时间也浪费不了多少。。。 
 

#include<iostream>
#include<cstdio>
using namespace std;
int main() {
  // freopen("input.txt", "r", stdin);
  double a,A[1001] = { 0 };
  int n,k, count = 0;
  cin >> k;
  while (k--) { 
    cin >> n >> a;
    A[n] += a; 
  }
  cin >> k;
  while (k--) { 
    cin >> n >> a;
    A[n] += a; 
  }
  for (int i = 0;i <= 1000;i++) if (A[i] != 0) count++;
  cout << count;
  for (int i = 1000;i >= 0;i--) {
    if (A[i] != 0) printf(" %d %.1f", i, A[i]);
  }
  return 0;
}