pat甲级1002

1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

主要是细节，要注意有可能相加后某个项会被消掉，这时候总项数要减少，而且为零项不要输出。还有一个效率问题，在读取时就把可能要打印的项数的指数存到一个set（排异性），减少
扫描时间。

#include <iostream>
#include <cstring>
#include <memory>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <set>

using namespace std;

double a[1002];
int main(void)
{
  memset(a, 0, sizeof(a));
  set<int> s;

  for (int i = 0;i < 2;i++)
  {
    int k;
    cin >> k;
    for (int j = 0;j < k;j++)
    {
      int index;
      double n;
      cin >> index >> n;
      a[index] += n;
      s.insert(index);//保存
    }
  }
  
  
  vector<int> v;
  v.assign(s.begin(), s.end());//用set构造

  int cnt = 0;
  for (int i = v.size();i > 0;--i)
    if (a[v[i - 1]] != 0.0)cnt++;//判断剩下的k
  cout << cnt;
  for (int i = v.size();i > 0;--i)
    if (a[v[i - 1]] != 0.0)//这里也要判断--
      printf(" %d %.1lf", v[i - 1], a[v[i - 1]]);
  printf("
");

  return 0;
}