PAT --甲级1002（1002 A+B for Polynomials）

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1002 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:

$K$ N1N_1N​1​​ aN1a_{N_1}a​N​1​​​​ N2N_2N​2​​ aN2a_{N_2}a​N​2​​​​ ... NKN_KN​K​​ aNKa_{N_K}a​N​K​​​​

where $K$ is the number of nonzero terms in the polynomial, NiN_iN​i​​ and aNia_{N_i}a​N​i​​​​ ($i=1,2,\cdots ,K$) are the exponents and coefficients, respectively. It is given that $1\le K\le 10$，0≤NK<⋯<N2<N1≤10000 \le N_K < \cdots < N_2 < N_1 \le 10000≤N​K​​<⋯<N​2​​<N​1​​≤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

给两个多项式，分别求其相加的结果。输入格式：K表示该多项式的项数，N表示幂，ai表示多项式的系数

思路：水题，细节很重要，注意保留1位小数；然后系数为0的项要忽略，无论是相加后为0还是输入时就为0都不应该输出。其实这题桶排好一点，贴上打了n个补丁的代码。噢对，如果幂不是整数的话要使用map，这题系数为整数。

#include <bits/stdc++.h>
using namespace std;

typedef pair<int, double > T;
T l1[20], l2[20];
int K1, K2;
vector<T> tab;
int vis1[1001], vis2[1001];

bool MyCompare (T a, T b) {
  return a.first > b.first;
}

int main() {
  memset(vis1, -1, sizeof(vis1));
  memset(vis2, -1, sizeof(vis2));
  cin >> K1;
  for (int i = 0; i < K1; i++) {
    cin >> l1[i].first >> l1[i].second;
    if (vis1[l1[i].first] != -1) {
      l1[vis1[l1[i].first]].second += l1[i].second;
      i--;
      K1--;
    } else {
      vis1[l1[i].first] = i;
    }
  }
  cin >> K2;
  for (int i = 0; i < K2; i++) {
    cin >> l2[i].first >> l2[i].second;
    if (vis2[l2[i].first] != -1) {
      l2[vis2[l2[i].first]].second += l2[i].second;
      i--;
      K2--;
    } else {
      vis2[l2[i].first] = i;
    }
  }
  sort(l1, l1 + K1, MyCompare);
  sort(l2, l2 + K2, MyCompare);
  int i = 0, j = 0;
  while (i < K1 && j < K2) {
    if (l1[i].second == 0) {
      i++;
      continue;
    }
    if (l2[j].second == 0) {
      j++;
      continue;
    }
    if (l1[i].first == l2[j].first) {
      if (l1[i].second + l2[j].second != 0) {
        tab.push_back(make_pair(l1[i].first, l1[i].second + l2[j].second));
      }
      i++;
      j++;
    } else if (l1[i].first > l2[j].first) {
      tab.push_back(make_pair(l1[i].first, l1[i].second));
      i++;
    } else {
      tab.push_back(make_pair(l2[j].first, l2[j].second));
      j++;
    }
  }
  for (; i < K1; i++) {
    tab.push_back(make_pair(l1[i].first, l1[i].second));
  }
  for (; j < K2; j++) {
    tab.push_back(make_pair(l2[j].first, l2[j].second));
  }
  cout << tab.size();
  for (vector<T>::iterator it = tab.begin(); it != tab.end(); it++) {
    cout << " " << it->first << " " << setiosflags(ios::fixed) << setprecision(1) << it->second;
  }
  cout << endl;
  return 0;
}