ＰＡＴ甲　1002. A+B for Polynomials (25)

最新推荐文章于 2024-03-12 00:56:10 发布

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最新推荐文章于 2024-03-12 00:56:10 发布

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本文介绍了一种解决多项式加法问题的算法实现，输入为两个多项式，输出为这两个多项式的和。算法通过读取系数和指数来构建多项式，并进行相加运算，最后输出结果。

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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

#inc<span style="font-family: "Droid Sans Mono", Consolas, "Courier New", monospace; font-size: 1em; line-height: 1.5;">lu</span><span style="font-family: "Droid Sans Mono", Consolas, "Courier New", monospace; font-size: 1em; line-height: 1.5;">de<iostream></span>
#include<cstring>
using namespace std;

int main()
{
  int i, k1, k2, j,max,count;
    double a[5][1005], b[1005];
  while (cin >> k1)
  {
    max = 0;
    memset(a, 0, sizeof(a));
    for (i = 0; i < k1; i++)
    {
      cin >> j;
      cin >> a[0][j];
      if (max < j)
        max = j;
    }
    cin >> k2;
    for (i = 0; i < k2; i++)
    {
      cin >> j;
      cin >> a[1][j];
      if (max < j)
        max = j;
    }
    memset(b, 0, sizeof(b));
    for (j = 0; j <= 1; j++)
    {

      for (i = 0; i <=max; i++)
      {
        b[i] += a[j][i];
      }
    }
    count = 0;
    for (i = 0; i <= max; i++)
    {
      if (b[i] != 0)
      {
        count++;
      }
    }
      cout << count;
      for (i = max; i >=0; i--)
      {
        if (b[i] != 0)
        {
          printf(" %d %.1f", i, b[i]);
        }
      }
      cout << endl;
    
  }
  return 0;
}