PAT 1002 A+B for Polynomials (25分)

题目

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≤N
K <⋯<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

题目解读

给出两个多项式，每个输入格式是非零项个数指数1 系数1 指数2 系数2
让计算两多项式的和，并按照指定格式输出 非零项个数指数1 系数1 指数2 系数2
要求顺序是指数从高到低。

思路解析

可以用一个结构体来保存每一项的指数和系数，然后在第二次输入时根据去找到相应的那一项，对其系数进行修改。
这样做既浪费存储空间也浪费时间，但一般都能想到，更好的做法是，用一个数组来取代整个结构体，每一项的指数作为数组的索引，系数作为值，这样在读入时，直接找到对应位置进行修改，对数组的访问是很快的。
之后一次遍历，统计出数组不为0的个数，就是非零项的个数；然后对数组从后往前输出每个非零项对应的下标和值，就是结果。

代码

#include <iostream>
using namespace std;
int main() {
  int a, b;
  cin >> a >> b;
  // 两数和转为字符串
  string s = to_string(a + b);
  // 得到有效长度
  int len = s.length();
  for (int i = 0; i < len; i++) {
    // 输出当前位
    cout << s[i];
    if (s[i] == '-')
      continue;
    // 标准化格式 -xx,123,999
    if ((i + 1) % 3 == len % 3 && i != len - 1)
      cout << ",";
  }
  return 0;
}