pat 1002 A+B for Polynomials (25 分)

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:

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 1K10,0NK​​<<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
 1 #include <map>
 2 #include <set>
 3 #include <queue>
 4 #include <cmath>
 5 #include <stack>
 6 #include <vector>
 7 #include <string>
 8 #include <cstdio>
 9 #include <cstring>
10 #include <climits>
11 #include <iostream>
12 #include <algorithm>
13 #define wzf ((1 + sqrt(5.0)) / 2.0)
14 #define INF 0x3f3f3f3f
15 #define eps 0.0000001
16 #define LL long long
17 using namespace std;
18 
19 const int MAXN = 1e3 + 10;
20 int cnt = 0, book[MAXN] = {0}, n, a;
21 double A[MAXN] = {0.0}, B[MAXN] = {0.0}, C[MAXN] = {0.0}, b;
22 
23 int main()
24 {
25     freopen("Date1.txt", "r", stdin);
26     scanf("%d", &n);
27     while (n --)
28     {
29         scanf("%d%lf", &a, &b);
30         A[a] = b;
31     }
32     scanf("%d", &n);
33     while (n --)
34     {
35         scanf("%d%lf", &a, &b);
36         B[a] += b;
37     }
38 
39     for (int i = 1005; i >= 0; -- i)
40     {
41         if (A[i] != 0 || B[i] != 0)
42             C[i] = A[i] + B[i];
43         if (C[i] != 0)
44             ++ cnt;
45     }
46     printf("%d", cnt);
47     for (int i = 1005; i >= 0; -- i)
48     {
49         if (C[i] != 0)
50             printf(" %d %.1f", i, C[i]);
51     }
52     printf("\n");
53     return 0;
54 }

 

posted @ 2018-09-15 17:59  GetcharZp  阅读(199)  评论(0编辑  收藏  举报